Question

14. Factor x4 + 3x2 - 28.O(x2-7)(x - 2)(x + 2)(x2 - 2)(x2 + 14)(x2 + 7)(x-2)(x + 2)O(x² + 4)(x² – 7)

Answer 1

We want to factor the expression;

$x^4+3x^2-28$

This can by the normal quadratic factorization because the expression even though a

quartic, is quadratic in x^2. Let's simplify it;

$(x^2)^2+3(x^2)-28$

Let us multiply both the constant term and the second power term to obtain

$-28(x^2)^2$

Let us think of numbers that multiply to give this number and also add to give,

$+3x^2$

The two numbers are;

$-4x^2and+7x^2$

Let's recast the equation then;

$\begin{gathered} (x^2)^2-4x^2+7x^2-28 \\ we\text{ can factorize each pair} \\ x^2(x^2-4)+7(x^2-4) \\ (x^2+7)(x^2-4) \end{gathered}$

We can factorize one of the terms further as it is a difference of two squares.

$(x^2+7)(x^2-4)=(x^2+7)(x-2)(x+2)$

Thus, our final answer is Eoption A