Answer:
[tex]x^4+3x^2-28=(x^2+7)(x-2)(x+2)[/tex]
Step-by-step explanation:
To factorize the expression, we can use a variable substitution. Let's say that z=x^2.
[tex]\begin{gathered} x^4+3x^2-28 \\ z^2+3z-28 \end{gathered}[/tex]
Then, to factorize this we need to factor in the form:
[tex](z+\text{?)(z}+\text{?)}[/tex]
The numbers that go in the blanks, have to:
*Add together to get 3
[tex]-4+7=3[/tex]
*Multiply together to get -28
[tex]-4\cdot7=-28[/tex]
So, we get:
[tex]z^2+3z-28=(z-4)(z+7)[/tex]
Substitute the equation z=x^2
[tex](x^2-4)(x^2+7)[/tex]
Factorizing the perfect square binomial:
[tex]x^4+3x^2-28=(x^2+7)(x-2)(x+2)[/tex]
