Answer:
![x^4+3x^2-28=(x^2+7)(x-2)(x+2)](https://img.qammunity.org/2023/formulas/mathematics/college/a4mwf29pzhyebkblik2wada4cirkhohuoy.png)
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}](https://img.qammunity.org/2023/formulas/mathematics/college/ixbxmekcrd9ft2md3ba56rxckta702fuev.png)
Then, to factorize this we need to factor in the form:
![(z+\text{?)(z}+\text{?)}](https://img.qammunity.org/2023/formulas/mathematics/college/9xi399u47qblubi0vuscy4y1kfrvchkicq.png)
The numbers that go in the blanks, have to:
*Add together to get 3
![-4+7=3](https://img.qammunity.org/2023/formulas/mathematics/college/57e0qfdzdreufhuuk8wc1pzfjejje6krje.png)
*Multiply together to get -28
![-4\cdot7=-28](https://img.qammunity.org/2023/formulas/mathematics/college/23h6h2zmpey8y8prv0beqcfbaqcdw8lizs.png)
So, we get:
![z^2+3z-28=(z-4)(z+7)](https://img.qammunity.org/2023/formulas/mathematics/college/rwoydxdgcywr9nsh3dfmnw1tx7021dn5u9.png)
Substitute the equation z=x^2
![(x^2-4)(x^2+7)](https://img.qammunity.org/2023/formulas/mathematics/college/j3739iwpa53k692x51o5eiqzi1b489gk41.png)
Factorizing the perfect square binomial:
![x^4+3x^2-28=(x^2+7)(x-2)(x+2)](https://img.qammunity.org/2023/formulas/mathematics/college/a4mwf29pzhyebkblik2wada4cirkhohuoy.png)