Question

ct the error in factoring the polynomial.
22.
$6x^2 - 7x - 3 = (3x - 3)(2x + 1)$

Explanation:

Step1: Verify the incorrect factorization

Expand $(3x-3)(2x+1)$:

$$\begin{align*} (3x-3)(2x+1)&=3x\cdot2x + 3x\cdot1 -3\cdot2x -3\cdot1\\ &=6x^2 + 3x -6x -3\\ &=6x^2 -3x -3 \end{align*}$$

This does not match $6x^2 -7x -3$, so the factorization is wrong.

Step2: Factor the polynomial correctly

Use the AC method: For $6x^2 -7x -3$, $a=6$, $b=-7$, $c=-3$.
Find two numbers that multiply to $a\cdot c=6\times(-3)=-18$ and add to $b=-7$. The numbers are $-9$ and $2$.
Rewrite the middle term:
$6x^2 -9x +2x -3$

Step3: Group and factor

Group terms: $(6x^2 -9x)+(2x -3)$
Factor out GCF from each group:
$3x(2x -3)+1(2x -3)$
Factor out $(2x-3)$:
$(3x+1)(2x-3)$

Step4: Verify correct factorization

Expand $(3x+1)(2x-3)$:

$$\begin{align*} (3x+1)(2x-3)&=3x\cdot2x +3x\cdot(-3)+1\cdot2x +1\cdot(-3)\\ &=6x^2 -9x +2x -3\\ &=6x^2 -7x -3 \end{align*}$$

This matches the original polynomial.

Answer:

The error is that the given factorization $(3x-3)(2x+1)$ expands to $6x^2 -3x -3$, not $6x^2 -7x -3$. The correct factorization is $(3x+1)(2x-3)$.