Question

6.4 trinomials where a do
(1 point)
factor:
$10x^4 + 3x^2 - 1 = \square$
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Explanation:

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

Original expression becomes \( 10y^2 + 3y - 1 \)

Step2: Factor quadratic in \( y \)

Find two numbers: \( 5 \times (-1) + 2 \times 1 = -5 + 2 = 3 \). So \( 10y^2 +5y -2y -1 = 5y(2y+1)-1(2y+1)=(5y-1)(2y+1) \)

Step3: Replace \( y \) with \( x^2 \)

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

Answer:

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