Question

factor using polynomial division
the polynomial $p(x) = 5x^3 - 9x^2 - 6x + 8$ has a known factor of $(x + 1)$.
rewrite $p(x)$ as a product of linear factors.
$p(x) = \square$

Explanation:

Step1: Divide \( p(x) \) by \( (x + 1) \)

We use polynomial long division to divide \( 5x^3 - 9x^2 - 6x + 8 \) by \( x + 1 \).
The first term of the quotient is \( \frac{5x^3}{x}=5x^2 \). Multiply \( (x + 1) \) by \( 5x^2 \) to get \( 5x^3+5x^2 \). Subtract this from \( 5x^3 - 9x^2 - 6x + 8 \):
\( (5x^3 - 9x^2 - 6x + 8)-(5x^3+5x^2)=-14x^2-6x + 8 \)
The next term of the quotient is \( \frac{-14x^2}{x}=-14x \). Multiply \( (x + 1) \) by \( -14x \) to get \( -14x^2-14x \). Subtract this from \( -14x^2-6x + 8 \):
\( (-14x^2-6x + 8)-(-14x^2-14x)=8x + 8 \)
The next term of the quotient is \( \frac{8x}{x}=8 \). Multiply \( (x + 1) \) by \( 8 \) to get \( 8x + 8 \). Subtract this from \( 8x + 8 \):
\( (8x + 8)-(8x + 8)=0 \)
So, \( 5x^3 - 9x^2 - 6x + 8=(x + 1)(5x^2-14x + 8) \)

Step2: Factor the quadratic \( 5x^2-14x + 8 \)

We need to find two numbers that multiply to \( 5\times8 = 40 \) and add up to \( -14 \). The numbers are \( -10 \) and \( -4 \).
Rewrite the middle term: \( 5x^2-10x-4x + 8 \)
Factor by grouping:
\( 5x(x - 2)-4(x - 2)=(5x - 4)(x - 2) \)

Step3: Write \( p(x) \) as product of linear factors

Since \( p(x)=(x + 1)(5x^2-14x + 8) \) and \( 5x^2-14x + 8=(5x - 4)(x - 2) \), we have:
\( p(x)=(x + 1)(5x - 4)(x - 2) \)

Answer:

\( (x + 1)(5x - 4)(x - 2) \)