Question

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

Answer 1

Answer:

$x^4+3x^2-28=(x^2+7)(x-2)(x+2)$

Step-by-step explanation:

To factorize the expression, we can use a variable substitution. Let's say that z=x^2.

$\begin{gathered} x^4+3x^2-28 \\ z^2+3z-28 \end{gathered}$

Then, to factorize this we need to factor in the form:

$(z+\text{?)(z}+\text{?)}$

The numbers that go in the blanks, have to:

*Add together to get 3

$-4+7=3$

*Multiply together to get -28

$-4\cdot7=-28$

So, we get:

$z^2+3z-28=(z-4)(z+7)$

Substitute the equation z=x^2

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

Factorizing the perfect square binomial:

$x^4+3x^2-28=(x^2+7)(x-2)(x+2)$