Polynomial identities and coefficient comparison | Year 12

01 / LEARN

The idea, step by step

An identity holds for every permitted input

An equation may be true only at particular values of $x$ . A polynomial identity is true for every real $x$ . Two real polynomials are identical precisely when their coefficients of every power match. Write missing terms with zero coefficients before comparing.

Use the identity symbol $\equiv$ when stressing this distinction. The claim $x^{2} \equiv x$ is false even though $x = 0$ and $x = 1$ both satisfy the associated equation. A few agreeing values do not automatically establish a polynomial identity.

Compare coefficients or choose strategic values

Expand the right-hand side, collect like powers, and equate its coefficients to those on the left. This produces ordinary equations for the unknown constants. When a factor vanishes at a convenient input, substitution can determine a constant with less expansion.

If the degree of the difference is known to be at most $n$ , agreement at $n + 1$ distinct inputs proves identity: a non-zero degree- $n$ polynomial cannot have more than $n$ distinct zeros. Without the degree bound, finite checking is not a proof.

A common trap

Checking an identity at one input proves it for all inputs.

Instead: One matching value is only a check. Compare coefficients, or use a justified degree bound with enough distinct inputs.

02 / WORKED EXAMPLES

See the thinking

Example 1

Comparing coefficients

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

Expand to $a x^{3} + (b + 2 a) x^{2} + (c + 2 b) x + 2 c$ .
From $x^{3}$ , $a = 2$ . From $x^{2}$ , $b + 4 = 7$ , so $b = 3$ . The constant gives $2 c = 2$ , so $c = 1$ .
The $x$ coefficient checks: $1 + 2 (3) = 7$ .

a = 2, b = 3, c = 1

Example 2

A useful input first

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

Set $x = 1$ : $8 = 3 B$ , so $B = \frac{8}{3}$ .
Set $x = - 2$ : $- 1 = - 3 A$ , so $A = \frac{1}{3}$ .
Check $A + B = 3$ and $- A + 2 B = 5$ .

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

Check your understanding

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

Show answer

$a = 3, b = - 2$ .

The coefficients of corresponding powers must agree.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF 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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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.

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