TMUA: Polynomials, Factor and Remainder Theorems

Polynomials: expanding, dividing and the Factor and Remainder Theorems.

Section 1 Part 1 · MM1 Algebra and functions · spec MM1.6
Papers 1 and 2
8 practice questions
No calculator

Practise polynomials, factor and remainder theorems →Timed TMUA paper

What the specification covers

Expanding and collecting terms; coefficient extraction
Division by linear and quadratic divisors
Factor and Remainder Theorems to find unknown coefficients

Key ideas

Remainder Theorem: the remainder when \(p(x)\) is divided by \(x-a\) is \(p(a)\).
Factor Theorem: \(x-a\) is a factor of \(p(x)\) exactly when \(p(a)=0\). For \(ax-b\), use \(p\!\left(\tfrac ba\right)\).
Dividing by a quadratic leaves a linear remainder \(rx+s\); find \(r\) and \(s\) by substituting the roots of the divisor.
Two unknown coefficients need two conditions: set up two linear equations and solve.

Common mistakes

For a factor \((x+2)\) substitute \(x=-2\), not \(x=2\).
Collecting coefficients: check the constant term and the leading term first, as quick sanity checks.

Exam tip

Substituting a root of the divisor is almost always quicker than long division.

Worked example

The polynomial \(p(x)=x^3+ax^2+bx-6\) has \((x+2)\) as a factor, and leaves remainder \(-4\) when divided by \((x-1)\). What is the value of \(b\)?

\(-5\)
\(\frac{5}{3}\)
\(-\frac{5}{3}\)
\(\frac{8}{3}\)
\(-\frac{8}{3}\)

Answer: C: \(-\frac{5}{3}\)

Remainder Theorem: \(p(1)=1+a+b-6=-4\), so \(a+b=1\). Factor Theorem: \(p(-2)=-8+4a-2b-6=0\), so \(2a-b=7\). Adding, \(3a=8\), \(a=\tfrac{8}{3}\), and \(b=1-\tfrac{8}{3}=-\tfrac{5}{3}\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

What is the remainder when \(x^5-3x^3+2x-1\) is divided by \(x^2-1\)?

\(2x-1\)
\(1-2x\)
\(0\)
\(-1\)
\(1\)

Show the answer and solution

Answer: D: \(-1\)

Write \(x^5-3x^3+2x-1=(x^2-1)q(x)+rx+s\). At \(x=1\): \(1-3+2-1=-1=r+s\). At \(x=-1\): \(-1+3-2-1=-1=-r+s\). So \(r=0\), \(s=-1\): the remainder is \(-1\).

Question 2

The polynomial \(p(x)=x^3+ax^2+bx-4\) has \((x+1)\) as a factor, and leaves remainder \(44\) when divided by \((x-3)\). What is the value of \(a\)?

\(-3\)
\(3\)
\(4\)
\(-2\)
\(1\)

Show the answer and solution

Answer: B: \(3\)

Factor Theorem: \(p(-1)=0\). Remainder Theorem: \(p(3)=44\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=3\), \(b=-2\).

Question 3

The polynomial \(p(x)=x^3+ax^2+bx+6\) has \((x-3)\) as a factor, and leaves remainder \(28\) when divided by \((x+1)\). What is the value of \(a\)?

\(3\)
\(1\)
\(-3\)
\(-20\)
\(4\)

Show the answer and solution

Answer: A: \(3\)

Factor Theorem: \(p(3)=0\). Remainder Theorem: \(p(-1)=28\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=3\), \(b=-20\).

Question 4

The polynomial \(p(x)=x^3+ax^2+bx-6\) has \((x+1)\) as a factor, and leaves remainder \(-48\) when divided by \((x-3)\). What is the value of \(a\)?

\(-6\)
\(-11\)
\(-3\)
\(-4\)
\(4\)

Show the answer and solution

Answer: D: \(-4\)

Factor Theorem: \(p(-1)=0\). Remainder Theorem: \(p(3)=-48\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=-4\), \(b=-11\).

Question 5

The polynomial \(p(x)=x^3+ax^2+bx+8\) has \((x-2)\) as a factor, and leaves remainder \(20\) when divided by \((x-3)\). What is the value of \(a\)?

\(1\)
\(-3\)
\(3\)
\(-14\)
\(4\)

Show the answer and solution

Answer: C: \(3\)

Factor Theorem: \(p(2)=0\). Remainder Theorem: \(p(3)=20\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=3\), \(b=-14\).

Question 6

The polynomial \(p(x)=x^3+ax^2+bx+8\) has \((x+1)\) as a factor, and leaves remainder \(20\) when divided by \((x-3)\). What is the value of \(a\)?

\(-2\)
\(-5\)
\(4\)
\(-3\)
\(3\)

Show the answer and solution

Answer: D: \(-3\)

Factor Theorem: \(p(-1)=0\). Remainder Theorem: \(p(3)=20\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=-3\), \(b=4\).

Question 7

The polynomial \(p(x)=x^3+ax^2+bx-8\) has \((x+2)\) as a factor, and leaves remainder \(-48\) when divided by \((x-2)\). What is the value of \(a\)?

\(4\)
\(-6\)
\(-3\)
\(-16\)
\(-4\)

Show the answer and solution

Answer: E: \(-4\)

Factor Theorem: \(p(-2)=0\). Remainder Theorem: \(p(2)=-48\). These are two linear equations in \(a\) and \(b\); solving them gives \(a=-4\), \(b=-16\).

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