what is the factor of x^4-x^2+16 and x^4-7x^2y^2+9y^4

Replace x^2 with b. This would make x^2 turn into b and x^4 turn into b^2.

Explain Bot Explain Bot answered
9 months ago

To find the factors of the given polynomials, we can try factoring by substitution. By replacing x^2 with b, we can rewrite the polynomials as follows:

First polynomial: b^2 - x^2 + 16
Second polynomial: b^2 - 7x^2y^2 + 9y^4

Now, let's factor the polynomials using the substitution:

First polynomial:
b^2 - x^2 + 16
= b^2 - (b^2) + 16
= 16

So, the first polynomial, x^4 - x^2 + 16, can be factored as 16.

Second polynomial:
b^2 - 7x^2y^2 + 9y^4
= b^2 - 7(b^2)(y^2) + 9y^4
= (b - 3y^2)(b - 3y^2)

Therefore, the second polynomial, x^4 - 7x^2y^2 + 9y^4, can be factored as (x^2 - 3y^2)^2.

In summary, the factors of the given polynomials are:
- For x^4 - x^2 + 16: 16
- For x^4 - 7x^2y^2 + 9y^4: (x^2 - 3y^2)^2