Question

$-7x^4 - 13x^2 + 6$

Explanation:

Step1: Let \( y = x^2 \)

Rewrite the polynomial: \( -7y^2 - 13y + 6 \)

Step2: Factor out -1

\( - (7y^2 + 13y - 6) \)

Step3: Factor the quadratic

Find two numbers: \( 14 \) and \( -3 \) (since \( 14 \times (-3) = -42 \), \( 14 + (-3) = 11 \)? Wait, no, correct for \( 7y^2 + 13y - 6 \): multiply \( 7 \times (-6) = -42 \), find two numbers that multiply to -42 and add to 13: 14 and -3. So split middle term: \( 7y^2 + 14y - 3y - 6 \)

Step4: Group and factor

\( (7y^2 + 14y) + (-3y - 6) = 7y(y + 2) - 3(y + 2) = (7y - 3)(y + 2) \)

Step5: Substitute back \( y = x^2 \) and apply -1

\( - (7x^2 - 3)(x^2 + 2) = (-7x^2 + 3)(x^2 + 2) \) or \( (3 - 7x^2)(x^2 + 2) \)

Answer:

\( (-7x^2 + 3)(x^2 + 2) \) (or equivalent factored form)