{"id":65344,"date":"2026-08-05T15:31:16","date_gmt":"2026-08-05T14:31:16","guid":{"rendered":"https:\/\/blog-admin.thethinkacademy.com\/?p=65344"},"modified":"2026-08-05T15:31:18","modified_gmt":"2026-08-05T14:31:18","slug":"polynomial-remainder-theorem-guide","status":"publish","type":"post","link":"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/08\/05\/polynomial-remainder-theorem-guide\/","title":{"rendered":"Polynomial Remainder Theorem Explained: Definition, Proof and Examples"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The polynomial remainder theorem is one of the most elegant shortcuts in algebra. Instead of performing polynomial long division to find the remainder, the theorem lets you find it by evaluating the polynomial at a single point. It is a prerequisite for the Factor Theorem, a foundational tool in MCR3U Unit 2, and appears in both curriculum assessments and mathematics competitions. This guide explains the theorem, proves it, walks through worked examples, and connects it to the broader polynomial toolkit.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">A small sign error can turn a straightforward remainder question into lost marks. Our free math assessment identifies how confidently your child applies polynomial concepts\u2014and where they may need more support before their next test.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em><a href=\"https:\/\/www.thinkacademy.ca\/free-assessment?source_id=6721&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\">Start the free math assessment \u2192<\/a><\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">What Is the Polynomial Remainder Theorem?<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The polynomial remainder theorem<\/strong> states:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">When a polynomial f(x) is divided by a linear divisor (x \u2212 a), the <strong>remainder<\/strong> is equal to <strong>f(a)<\/strong>.<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mi>f<\/mi><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo stretchy=\"false\">)<\/mo><mo>=<\/mo><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo>\u2212<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>\u22c5<\/mo><mi>q<\/mi><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo stretchy=\"false\">)<\/mo><mo>+<\/mo><mi>r<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">f(x) = (x &#8211; a) \\cdot q(x) + r<\/annotation><\/semantics><\/math>f(x)=(x\u2212a)\u22c5q(x)+r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">where q(x) is the quotient polynomial and r is the remainder. Substituting x = a:<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mi>f<\/mi><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>=<\/mo><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo>\u2212<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>\u22c5<\/mo><mi>q<\/mi><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>+<\/mo><mi>r<\/mi><mo>=<\/mo><mn>0<\/mn><mo>\u22c5<\/mo><mi>q<\/mi><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>+<\/mo><mi>r<\/mi><mo>=<\/mo><mi>r<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">f(a) = (a &#8211; a) \\cdot q(a) + r = 0 \\cdot q(a) + r = r<\/annotation><\/semantics><\/math>f(a)=(a\u2212a)\u22c5q(a)+r=0\u22c5q(a)+r=r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So <strong>r = f(a)<\/strong> \u2014 the remainder is simply the value of the polynomial at x = a.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>In plain English:<\/strong> to find the remainder when f(x) is divided by (x \u2212 a), substitute x = a directly into f(x). No long division required.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Proof of the Polynomial Remainder Theorem<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">The proof is short and relies only on the definition of polynomial division.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Claim:<\/strong> When f(x) is divided by (x \u2212 a), the remainder is f(a).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Proof:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By the division algorithm for polynomials, we can always write:<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mi>f<\/mi><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo stretchy=\"false\">)<\/mo><mo>=<\/mo><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo>\u2212<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>\u22c5<\/mo><mi>q<\/mi><mo stretchy=\"false\">(<\/mo><mi>x<\/mi><mo stretchy=\"false\">)<\/mo><mo>+<\/mo><mi>r<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">f(x) = (x &#8211; a) \\cdot q(x) + r<\/annotation><\/semantics><\/math>f(x)=(x\u2212a)\u22c5q(x)+r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">where q(x) is a polynomial (the quotient) and r is a constant (the remainder, since the divisor is linear and of degree 1, the remainder has degree less than 1 \u2014 so it is a constant).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substitute x = a:<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mi>f<\/mi><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>=<\/mo><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo>\u2212<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>\u22c5<\/mo><mi>q<\/mi><mo stretchy=\"false\">(<\/mo><mi>a<\/mi><mo stretchy=\"false\">)<\/mo><mo>+<\/mo><mi>r<\/mi><mo>=<\/mo><mn>0<\/mn><mo>+<\/mo><mi>r<\/mi><mo>=<\/mo><mi>r<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">f(a) = (a &#8211; a) \\cdot q(a) + r = 0 + r = r<\/annotation><\/semantics><\/math>f(a)=(a\u2212a)\u22c5q(a)+r=0+r=r<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore r = f(a). \u25a1<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This proof is short enough to reproduce from memory \u2014 and knowing it makes the theorem feel logical rather than arbitrary.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">The Polynomial Remainder Theorem vs The Factor Theorem<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">These two theorems are closely related. Understanding the distinction is essential for MCR3U.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Polynomial Remainder Theorem<\/th><th>Factor Theorem<\/th><\/tr><\/thead><tbody><tr><td><strong>What it says<\/strong><\/td><td>The remainder when f(x) is divided by (x \u2212 a) is f(a)<\/td><td>(x \u2212 a) is a factor of f(x) if and only if f(a) = 0<\/td><\/tr><tr><td><strong>When remainder = 0<\/strong><\/td><td>f(a) = 0<\/td><td>(x \u2212 a) divides f(x) exactly<\/td><\/tr><tr><td><strong>Relationship<\/strong><\/td><td>General case<\/td><td>Special case where remainder = 0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The Factor Theorem is a special case of the polynomial remainder theorem.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If f(a) = 0 (remainder is zero), then (x \u2212 a) is a factor \u2014 division leaves no remainder. This is the Factor Theorem.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If f(a) \u2260 0, the polynomial remainder theorem gives you the exact remainder without division.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/www.thinkacademy.ca\/free-assessment?source_id=6721&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"409\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-1024x409.png\" alt=\"polynomial remainder theorem cta can your child avoid this common mistake try the free math assessment\" class=\"wp-image-65346\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-1024x409.png 1024w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-300x120.png 300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-768x307.png 768w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-1536x614.png 1536w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1-1300x520.png 1300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_20_33-PM-1.png 1983w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Worked Examples<\/h2>\n\n\n\n<h3 class=\"wp-block-heading\">Example 1 \u2014 Finding a Remainder (MCR3U Level)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Find the remainder when f(x) = x\u00b3 \u2212 4x\u00b2 + 3x \u2212 7 is divided by (x \u2212 2).<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By the polynomial remainder theorem: remainder = f(2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">f(2) = (2)\u00b3 \u2212 4(2)\u00b2 + 3(2) \u2212 7 = 8 \u2212 16 + 6 \u2212 7 = <strong>\u22129<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The remainder is \u22129. No long division required.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Verify with long division (optional):<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">x\u00b3 \u2212 4x\u00b2 + 3x \u2212 7 \u00f7 (x \u2212 2) = x\u00b2 \u2212 2x \u2212 1 with remainder \u22129 \u2713<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example 2 \u2014 Divisor of the Form (x + a)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Find the remainder when f(x) = 2x\u00b3 + x\u00b2 \u2212 5x + 1 is divided by (x + 3).<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Write (x + 3) = (x \u2212 (\u22123)), so a = \u22123.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Remainder = f(\u22123):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">f(\u22123) = 2(\u22123)\u00b3 + (\u22123)\u00b2 \u2212 5(\u22123) + 1 = 2(\u221227) + 9 + 15 + 1 = \u221254 + 25 = <strong>\u221229<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The remainder is \u221229.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example 3 \u2014 Divisor (ax \u2212 b): Extension<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Find the remainder when f(x) = 3x\u00b3 \u2212 x\u00b2 + 2x \u2212 4 is divided by (3x \u2212 1).<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Write 3x \u2212 1 = 3(x \u2212 1\/3), so set x = 1\/3.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Remainder = f(1\/3):<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">f(1\/3) = 3(1\/3)\u00b3 \u2212 (1\/3)\u00b2 + 2(1\/3) \u2212 4 = 3(1\/27) \u2212 1\/9 + 2\/3 \u2212 4 = 1\/9 \u2212 1\/9 + 2\/3 \u2212 4 = 2\/3 \u2212 4 = <strong>\u221210\/3<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>Note: When dividing by a non-monic linear divisor (ax \u2212 b), the remainder is still f(b\/a). The quotient changes, but the remainder from the polynomial remainder theorem is the same.<\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example 4 \u2014 Working Backwards: Finding an Unknown Coefficient<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The polynomial f(x) = x\u00b3 + kx\u00b2 \u2212 2x + 5 has remainder 11 when divided by (x \u2212 2). Find k.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By the polynomial remainder theorem: f(2) = 11<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">f(2) = (2)\u00b3 + k(2)\u00b2 \u2212 2(2) + 5 = 8 + 4k \u2212 4 + 5 = 9 + 4k<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Set equal to 11: 9 + 4k = 11 4k = 2 <strong>k = 1\/2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This type of problem \u2014 finding a coefficient from a given remainder \u2014 is a standard MCR3U exam question type and appears regularly in MHF4U assessments.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example 5 \u2014 Two Conditions, System of Equations (MHF4U Level)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The polynomial f(x) = x\u00b3 + ax\u00b2 + bx \u2212 3 has remainder 5 when divided by (x \u2212 2) and remainder \u22127 when divided by (x + 1). Find a and b.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From (x \u2212 2): f(2) = 5 8 + 4a + 2b \u2212 3 = 5 4a + 2b = 0 <strong>2a + b = 0<\/strong> &#8230; (1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From (x + 1): f(\u22121) = \u22127 \u22121 + a \u2212 b \u2212 3 = \u22127 a \u2212 b = \u22123 <strong>a \u2212 b = \u22123<\/strong> &#8230; (2)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Add (1) and (2): 3a = \u22123 \u2192 <strong>a = \u22121<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substitute into (1): 2(\u22121) + b = 0 \u2192 <strong>b = 2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">f(x) = x\u00b3 \u2212 x\u00b2 + 2x \u2212 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><em>Verify: f(2) = 8 \u2212 4 + 4 \u2212 3 = 5 \u2713 and f(\u22121) = \u22121 \u2212 1 \u2212 2 \u2212 3 = \u22127 \u2713<\/em><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This two-condition system is the most algebraically demanding polynomial remainder theorem problem type at the MCR3U\/MHF4U level.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/www.thinkacademy.ca\/free-assessment?source_id=6721&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"409\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-1024x409.png\" alt=\"\" class=\"wp-image-65347\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-1024x409.png 1024w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-300x120.png 300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-768x307.png 768w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-1536x614.png 1536w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1-1300x520.png 1300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_20-PM-1.png 1983w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example 6 \u2014 Connecting to the Factor Theorem<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Determine whether (x \u2212 3) is a factor of f(x) = x\u00b3 \u2212 2x\u00b2 \u2212 5x + 6.<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By the polynomial remainder theorem: f(3) = (3)\u00b3 \u2212 2(3)\u00b2 \u2212 5(3) + 6 = 27 \u2212 18 \u2212 15 + 6 = <strong>0<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since f(3) = 0, the remainder is 0, so (x \u2212 3) <strong>is<\/strong> a factor of f(x).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Factor: f(x) \u00f7 (x \u2212 3) = x\u00b2 + x \u2212 2 = (x + 2)(x \u2212 1)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>f(x) = (x \u2212 3)(x + 2)(x \u2212 1)<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Where the Polynomial Remainder Theorem Appears in Contests and Curriculum<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>MCR3U (Unit 2: Polynomial Functions):<\/strong> The polynomial remainder theorem is explicitly taught and assessed in Ontario&#8217;s Grade 11 curriculum. Unit tests and final exams include direct remainder calculations (Examples 1\u20132), working backwards to find a coefficient (Example 4), and combined use with the Factor Theorem (Example 6). Students who can apply the theorem quickly \u2014 finding f(a) by substitution rather than long division \u2014 complete polynomial questions significantly faster. See our <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/mcr3u-grade-11-functions-ontario\/\">MCR3U complete guide<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>MHF4U (Polynomial and Rational Functions Unit):<\/strong> MHF4U builds on the polynomial remainder theorem with higher-degree polynomials, more complex coefficient-finding problems, and the two-condition system type (Example 5). Students are expected to apply the theorem fluently without re-deriving it. See our <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/07\/09\/mhf4u-advanced-functions\/\">MHF4U Advanced Functions guide<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Euclid Contest (CEMC, Grade 12):<\/strong> Polynomial algebra appears regularly in Euclid Part B \u2014 typically in the form of a multi-step problem where the polynomial remainder theorem is one tool among several (combined with the rational root theorem, the Factor Theorem, or system-solving). A student who is slow at applying the theorem by hand will lose time on these problems. See our <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/05\/18\/euclid-math-contest-preparation-guide-canada\/\">Euclid math contest guide<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>AMC 10 and Cayley Contest:<\/strong> Polynomial problems at the AMC 10 and Cayley level occasionally involve remainder-type questions, particularly asking for the remainder when a polynomial is evaluated at a specific value. Knowing the polynomial remainder theorem allows direct evaluation where other students attempt full division. See our <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/05\/19\/cayley-math-contest-complete-guide-canada\/\">Cayley math contest guide<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">For the full competition landscape, see our <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/06\/23\/math-competition-canada\/\">math competitions in Canada guide<\/a>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">The Polynomial Remainder Theorem and Synthetic Division<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Synthetic division and the polynomial remainder theorem are complementary tools:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>polynomial remainder theorem<\/strong> gives the remainder instantly via substitution \u2014 f(a) \u2014 without any division<\/li>\n\n\n\n<li><strong>Synthetic division<\/strong> performs the actual division and yields both the quotient polynomial and the remainder simultaneously<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">When you only need the remainder, use the polynomial remainder theorem \u2014 it is faster.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">When you need both the remainder and the quotient (e.g., to continue factoring), use synthetic division \u2014 which also confirms the remainder as its final entry.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Example: f(x) = x\u00b3 \u2212 4x\u00b2 + 3x \u2212 7 divided by (x \u2212 2)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Polynomial remainder theorem: f(2) = 8 \u2212 16 + 6 \u2212 7 = \u22129. Done.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Synthetic division gives: quotient x\u00b2 \u2212 2x \u2212 1, remainder \u22129.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Both confirm the remainder is \u22129. For factoring purposes, synthetic division is needed for the quotient. For a remainder-only question, the polynomial remainder theorem is the efficient choice.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Common Mistakes<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mistake 1: Substituting the wrong value.<\/strong> When dividing by (x \u2212 a), substitute x = <strong>+a<\/strong> (not \u2212a). When dividing by (x + 3) = (x \u2212 (\u22123)), substitute x = <strong>\u22123<\/strong>. Students who substitute the wrong sign get the wrong remainder every time.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mistake 2: Confusing the remainder with the quotient.<\/strong> The polynomial remainder theorem gives the remainder \u2014 the constant left over. It does not give the quotient polynomial. If the question asks for the quotient, synthetic division is required.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mistake 3: Using the theorem with non-linear divisors.<\/strong> The theorem applies to division by linear factors (x \u2212 a) only. Dividing by x\u00b2 \u2212 1 or any non-linear expression requires full polynomial long division.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mistake 4: Arithmetic errors in evaluating f(a).<\/strong> The calculation of f(a) \u2014 especially for cubic and quartic polynomials with multiple terms \u2014 is where most errors occur. Write each term substitution on a separate line; do not try to compute f(a) in one mental step.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Mistake 5: Thinking remainder = 0 means the theorem doesn&#8217;t apply.<\/strong> If f(a) = 0, the remainder is simply 0 \u2014 the polynomial remainder theorem applies as always. The consequence (that (x \u2212 a) is a factor) is the Factor Theorem, which follows directly from the polynomial remainder theorem.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Practice Problems<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Set A \u2014 Direct application<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Find the remainder when f(x) is divided by the given divisor:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>f(x) = x\u00b3 + 2x\u00b2 \u2212 x + 4, divided by (x \u2212 3)<\/li>\n\n\n\n<li>f(x) = 2x\u00b3 \u2212 5x + 1, divided by (x + 2)<\/li>\n\n\n\n<li>f(x) = x\u2074 \u2212 3x\u00b3 + x \u2212 5, divided by (x \u2212 1)<\/li>\n\n\n\n<li>f(x) = 3x\u00b3 + x\u00b2 \u2212 7x + 2, divided by (x + 1)<\/li>\n\n\n\n<li>f(x) = x\u00b3 \u2212 8, divided by (x \u2212 2)<\/li>\n\n\n\n<li>f(x) = 5x\u2074 \u2212 2x\u00b3 + x \u2212 3, divided by (x + 1)<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Set B \u2014 Working backwards<\/strong><\/p>\n\n\n\n<ol start=\"7\" class=\"wp-block-list\">\n<li>f(x) = x\u00b3 + kx\u00b2 + 3x \u2212 5 has remainder 3 when divided by (x \u2212 2). Find k.<\/li>\n\n\n\n<li>f(x) = 2x\u00b3 \u2212 3x\u00b2 + mx + 4 has remainder \u22121 when divided by (x + 1). Find m.<\/li>\n\n\n\n<li>f(x) = x\u00b3 + ax + b has remainder 4 when divided by (x \u2212 1) and remainder \u22122 when divided by (x + 1). Find a and b.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Set C \u2014 Connecting to the Factor Theorem<\/strong><\/p>\n\n\n\n<ol start=\"10\" class=\"wp-block-list\">\n<li>Determine whether (x + 2) is a factor of f(x) = x\u00b3 + x\u00b2 \u2212 4x \u2212 4.<\/li>\n\n\n\n<li>For what value of k is (x \u2212 3) a factor of f(x) = x\u00b3 \u2212 kx\u00b2 + 2x \u2212 3?<\/li>\n\n\n\n<li>f(x) = x\u00b3 + 2x\u00b2 \u2212 5x + k. Find k so that (x \u2212 1) is a factor, then fully factor f(x).<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Answers:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Set A:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>f(3) = 27 + 18 \u2212 3 + 4 = <strong>46<\/strong><\/li>\n\n\n\n<li>f(\u22122) = \u221216 + 10 + 1 = <strong>\u22125<\/strong><\/li>\n\n\n\n<li>f(1) = 1 \u2212 3 + 1 \u2212 5 = <strong>\u22126<\/strong><\/li>\n\n\n\n<li>f(\u22121) = \u22123 + 1 + 7 + 2 = <strong>7<\/strong><\/li>\n\n\n\n<li>f(2) = 8 \u2212 8 = <strong>0<\/strong> (so (x\u22122) is a factor \u2014 f(x) = (x\u22122)(x\u00b2+2x+4))<\/li>\n\n\n\n<li>f(\u22121) = 5 + 2 \u2212 1 \u2212 3 = <strong>3<\/strong><\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Set B: 7) f(2) = 8 + 4k + 6 \u2212 5 = 9 + 4k = 3 \u2192 4k = \u22126 \u2192 <strong>k = \u22123\/2<\/strong> 8) f(\u22121) = \u22122 \u2212 3 \u2212 m + 4 = \u22121 \u2212 m = \u22121 \u2192 <strong>m = 0<\/strong> 9) f(1) = 1 + a + b = 4 \u2192 a + b = 3 &#8230; (1); f(\u22121) = \u22121 \u2212 a + b = \u22122 \u2192 \u2212a + b = \u22121 &#8230; (2); Add: 2b = 2 \u2192 b = 1; then a = 2. <strong>a = 2, b = 1<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Set C: 10) f(\u22122) = \u22128 + 4 + 8 \u2212 4 = <strong>0<\/strong> \u2713. Yes, (x+2) is a factor. f(x) = (x+2)(x\u00b2\u2212x\u22122) = (x+2)(x\u22122)(x+1) 11) f(3) = 0: 27 \u2212 9k + 6 \u2212 3 = 0 \u2192 30 \u2212 9k = 0 \u2192 <strong>k = 10\/3<\/strong> 12) f(1) = 0: 1 + 2 \u2212 5 + k = 0 \u2192 k = 2. f(x) = x\u00b3 + 2x\u00b2 \u2212 5x + 2. Divide by (x\u22121): quotient x\u00b2 + 3x \u2212 2. Discriminant = 9 + 8 = 17 \u2014 does not factor over integers. <strong>f(x) = (x\u22121)(x\u00b2 + 3x \u2212 2)<\/strong>. (Quadratic factor is irreducible over integers.)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Frequently Asked Questions<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>What is the polynomial remainder theorem?<\/strong> When a polynomial f(x) is divided by a linear factor (x \u2212 a), the remainder is f(a). This allows the remainder to be found by direct substitution rather than polynomial long division.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>What is the difference between the polynomial remainder theorem and the Factor Theorem?<\/strong> The polynomial remainder theorem gives the remainder for any division by (x \u2212 a). The Factor Theorem is the special case where the remainder is zero \u2014 meaning (x \u2212 a) divides f(x) exactly and is therefore a factor. Every application of the Factor Theorem is an application of the polynomial remainder theorem with result zero.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Does the polynomial remainder theorem work for non-linear divisors?<\/strong> No. The theorem applies only when dividing by a linear expression (x \u2212 a). For non-linear divisors, polynomial long division is required and the remainder is a polynomial, not a constant.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>How is the polynomial remainder theorem used to find unknown coefficients?<\/strong> If the remainder of f(x) \u00f7 (x \u2212 a) is known, set f(a) equal to that remainder value and solve for the unknown coefficient. This produces one equation per condition \u2014 two unknowns require two conditions and a system of equations (as in Example 5).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Is the polynomial remainder theorem on the MCR3U exam?<\/strong> Yes. It is explicitly part of the Ontario Grade 11 curriculum (Unit 2: Polynomial Functions) and is assessed on unit tests and final exams. The most commonly tested forms are direct remainder calculation and working backwards to find an unknown coefficient.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>What is the easiest way to remember the theorem?<\/strong> &#8220;Divide by (x \u2212 a), remainder is f(a).&#8221; The value that makes the divisor zero \u2014 x = a \u2014 is exactly the value you substitute into f(x) to get the remainder.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><em>See our related guides: <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/rational-root-theorem-guide\/\">rational root theorem guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/mcr3u-grade-11-functions-ontario\/\">MCR3U complete guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/07\/09\/mhf4u-advanced-functions\/\">MHF4U Advanced Functions guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/types-of-functions-in-math-guide\/\">types of function in math guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/composite-functions-maths-guide\/\">composite functions maths guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/05\/18\/euclid-math-contest-preparation-guide-canada\/\">Euclid math contest guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/05\/19\/cayley-math-contest-complete-guide-canada\/\">Cayley math contest guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/math-induction-proof-guide\/\">math induction proof guide<\/a> \u00b7 <a href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/06\/23\/math-competition-canada\/\">math competitions in Canada<\/a><\/em><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The polynomial remainder theorem is a one-line shortcut with exam marks attached. Make sure your child can use it cold.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><a href=\"https:\/\/www.thinkacademy.ca\/trialclass_en?source_id=6721&amp;source_type=9&amp;utm_medium=website&amp;utm_source=pc_blog\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"409\" src=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-1024x409.png\" alt=\"\" class=\"wp-image-65348\" srcset=\"https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-1024x409.png 1024w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-300x120.png 300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-768x307.png 768w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-1536x614.png 1536w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1-1300x520.png 1300w, https:\/\/blog-admin.thethinkacademy.com\/wp-content\/uploads\/2026\/08\/ChatGPT-Image-Aug-5-2026-03_26_53-PM-1.png 1983w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/a><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>The polynomial remainder theorem is one of the most elegant shortcuts in algebra. Instead of performing &hellip; <a title=\"Polynomial Remainder Theorem Explained: Definition, Proof and Examples\" class=\"hm-read-more\" href=\"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/08\/05\/polynomial-remainder-theorem-guide\/\"><span class=\"screen-reader-text\">Polynomial Remainder Theorem Explained: Definition, Proof and Examples<\/span>Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":65345,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[17232,17160,17135],"tags":[],"class_list":["post-65344","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-math-competitions","category-math-skills","category-study-tips-and-tools"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Polynomial Remainder Theorem: Definition, Proof, Examples<\/title>\n<meta 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