Polynomial identities and coefficient comparison | Year 12

01 / LEARN

Understand the idea.

An identity holds for every permitted input

An equation can hold at selected inputs; an identity holds for every input in its stated domain. Real polynomials are identical exactly when every corresponding coefficient agrees, including zero coefficients for missing powers.

The symbol $\equiv$ emphasises identity. For example, $x^{2} = x$ is an equation with solutions zero and one, but $x^{2} \equiv x$ is false. Matching a few inputs needs an additional degree argument to prove identity.

Compare coefficients or choose strategic values

Expand and collect powers before comparing coefficients. Each power gives an ordinary equation for the unknown constants. Alternatively, substitute an input that makes one factor zero to isolate a constant quickly.

A nonzero polynomial of degree at most $n$ has at most $n$ distinct roots. Therefore agreement at $n + 1$ distinct inputs proves identity when that degree bound is established. Otherwise numerical agreement remains evidence rather than proof.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Comparing coefficients

Find $a, b, c$ if $(x + 2) (a x^{2} + b x + c) \equiv 2 x^{3} + 7 x^{2} + 7 x + 2$ .

Worked solution

1
Expand and group by powers.
$\begin{aligned} (x + 2) (a x^{2} + b x + c) & = a x^{3} + (b + 2 a) x^{2} \\ + (c + 2 b) x + 2 c . \end{aligned}$
Corresponding coefficients can be compared only after like powers are collected.
2
Use the leading and quadratic coefficients.
$\begin{aligned} a & = 2, \\ b + 2 a & = 7 \Rightarrow b = 3. \end{aligned}$
An identity requires the $x^{3}$ and $x^{2}$ coefficients to agree.
3
Use the constant coefficient.
$2 c = 2 \Rightarrow c = 1.$
The constant terms must agree independently of $x$ .
4
Check the remaining coefficient.
$c + 2 b = 1 + 2 (3) = 7.$
All four coefficients now agree, proving the identity for $a = 2, b = 3, c = 1$ .

Answer

$a = 2, b = 3, c = 1$

EXAMPLE 02

A useful input first

Find $A, B$ in the identity $3 x + 5 \equiv A (x - 1) + B (x + 2)$ .

Worked solution

1
Eliminate the term containing $A$ .
$x = 1 : 3 (1) + 5 = A (0) + B (3), B = \frac{8}{3} .$
At $x = 1$ , the factor $x - 1$ vanishes.
2
Eliminate the term containing $B$ .
$x = - 2 : 3 (- 2) + 5 = A (- 3) + B (0), A = \frac{1}{3} .$
At $x = - 2$ , the factor $x + 2$ vanishes.
3
Verify both coefficients.
$\begin{aligned} A (x - 1) + B (x + 2) & = (A + B) x + (- A + 2 B), \\ A + B & = \frac{1}{3} + \frac{8}{3} = 3, \\ - A + 2 B & = - \frac{1}{3} + \frac{16}{3} = 5. \end{aligned}$
Matching the linear and constant coefficients proves that the found constants satisfy the identity.

Answer

$A = \frac{1}{3}, B = \frac{8}{3}$

03 / YOUR TURN

Try it yourself.

Check your understanding

If $a x^{2} + b x + 4 \equiv 3 x^{2} - 2 x + 4$ , find $a, b$ .

Show answer & reasoning

$a = 3, b = - 2$ .

Compare the $x^{2}$ coefficients to obtain $a = 3$ and the $x$ coefficients to obtain $b = - 2$ . The constant coefficients already agree.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find $a, b$ if $(x + 1) (x + a) \equiv x^{2} + 5 x + b$ .
2 marks
Find $a, b$ if $(2 x - 1) (x^{2} + a x + b) \equiv 2 x^{3} + 3 x^{2} - 4 x + 1$ .
2 marks
Find $A, B$ if $5 x - 1 \equiv A (x + 1) + B (x - 2)$ .
2 marks
A student claims $x^{3} - x \equiv 0$ because it is zero at $- 1, 0, 1$ . Explain the error.
2 marks
Find $a, b, c$ if $a x^{2} + b x + c \equiv {(x - 2)}^{2} + 3 (x + 1)$ .
2 marks
Two polynomials of degree at most $2$ agree at three distinct inputs. Must they be identical? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $a, b$ if $(x + 1) (x + a) \equiv x^{2} + 5 x + b$ .
2 marks
Expansion is $x^{2} + (a + 1) x + a$ . Compare the $x$ and constant coefficients.
$a = 4, b = 4$
Find $a, b$ if $(2 x - 1) (x^{2} + a x + b) \equiv 2 x^{3} + 3 x^{2} - 4 x + 1$ .
2 marks
$2 a - 1 = 3$ gives $a = 2$ ; $- b = 1$ gives $b = - 1$ ; then $2 b - a = - 4$ .
$a = 2, b = - 1$
Find $A, B$ if $5 x - 1 \equiv A (x + 1) + B (x - 2)$ .
2 marks
Set $x = 2$ to get $9 = 3 A$ ; set $x = - 1$ to get $- 6 = - 3 B$ .
$A = 3, B = 2$
A student claims $x^{3} - x \equiv 0$ because it is zero at $- 1, 0, 1$ . Explain the error.
2 marks
For example, at $x = 2$ the expression equals $6$ .
Three zeros are possible for a non-zero cubic; this is not an identity.
Find $a, b, c$ if $a x^{2} + b x + c \equiv {(x - 2)}^{2} + 3 (x + 1)$ .
2 marks
Expand: $x^{2} - 4 x + 4 + 3 x + 3 = x^{2} - x + 7$ .
$a = 1, b = - 1, c = 7$
Two polynomials of degree at most $2$ agree at three distinct inputs. Must they be identical? Explain.
3 marks
Their difference has degree at most $2$ and has three distinct zeros. Only the zero polynomial can do this.
Yes.

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Curriculum reference & lesson scope

SACE Board of South Australia

A recognisable real-polynomial chapter supporting Stage 2 Specialist polynomial factorisation and graph analysis; original lessons use current SACE scope and the school chapter structure requested by the user.

Official curriculum source ↗

In this lesson

Understand the idea.

An identity holds for every permitted input

Compare coefficients or choose strategic values

See the method in action.

Comparing coefficients

Expand and group by powers.

Use the leading and quadratic coefficients.

Use the constant coefficient.

Check the remaining coefficient.

A useful input first

Eliminate the term containing A.

Eliminate the term containing B.

Verify both coefficients.

Try it yourself.

Check your understanding

Keep practising

Eliminate the term containing $A$ .

Eliminate the term containing $B$ .