Factor by substitution:

$y^4-y^2-20$

We have an equation $y^4-y^2-20$.

We can see that the variable term of the middle term here is $y^2$and it's square is $y^4$.

If we substitute any variable of single power in place of the given variables, we will get an equation in the standard form and we can easily solve it by ac method.

Then by replacing the substituted variable with the original one will give us the answer.

We have $y^4-y^2-20$. We also know that ${\left(y^2\right)}^2=y^4$.

So, assuming$u=y^2$, we can write the equation as$u^2-u-20$.

We have the new equation as $u^2-u-20$.

We need two numbers whose sum comes out to be the product of first and last term, i.e., $-20$and the sum comes out to be the middle term, i.e., $-1$. As the product of the terms is negative and the sum is also negative, we need to take the sign of the larger value term as negative. The factors of $-20$and their sums are-

This shows that two numbers are$4,-5$.

We have two numbers as $4,-5$.

Splitting the middle term gives us,

$$\begin{gathered} =u^2-u-20 \\ =u^2+4u-5u-20 \\ =u(u+4)-5(u+4) \\ =(u+4)(u-5) \end{gathered}$$

Now replacing the variable$"u"$by$y^2$, we get the factors asrole="math" localid="1644584115294" $(y^2+4)(y^2-5)$.

We will check the solution by multiplying. If we get the same given equation, our calculations are right.

So,

$$\begin{gathered} =(y^2+4)(y^2-5) \\ =y^4-5y^2+4y^2-20 \\ =y^4-y^2-20 \end{gathered}$$