Practice Factoring General Polynomials

We will factor ‍$$y^{10}{+7}y^5{-8}$$ as $$(y^5+a)(y^5+b)$$.

We need to find numbers $$a$$ and ‍$$b$$ such that ‍$${a+b}={7}$$ and $${ab} ={-8}$$.

$$\begin{aligned}
(+8)+(-1)=7
\\
(+8)(-1)=-8
\end{aligned}$$

So the integers we are looking for are +8 and -1.

In conclusion, $$y^{10}+7y^5-8=(y^5+8)(y^5-1)$$.

Take a common factor of $$-5h^2$$..

$$-5h^4-15h^3-10h^2=-5h^2(h^2+3h+2)$$

Now let's factor $$V$$.

$$h^2+3h+2=(h+1)(h+2)$$

In conclusion, $$\begin{aligned}
-5h^4-15h^3-10h^2 =-5h^2(h^2+3h+2)
\\\\
&=-5h^2(h+1)(h+2)
\end{aligned}$$.

We will factor ‍$$x^2{-3}xy{-10}y^2$$ as $$(x+ay)(x+by)$$.

We need to find numbers $$a$$ and ‍$$b$$ such that ‍$${a+b}={-3}$$ and $${ab} ={-10}$$.

$$\begin{aligned}
(+2)+(-5)=-3
\\
(+2)(-5)=-10
\end{aligned}$$

So the integers we are looking for are +2 and -5.

In conclusion, $$x^2-3xy-10y^2=(x+2y)(x-5y)$$.

We will factor ‍$$n^8{-5}n^4{-6}$$ as $$(n^4+a)(n^4+b)$$.

We need to find numbers $$a$$ and ‍$$b$$ such that ‍$${a+b}={-5}$$ and $${ab} ={-6}$$.

$$\begin{aligned}
&(+1)+(-6)=-5
\\\\
&(+1)(-6)=-6
\end{aligned}$$

So the integers we are looking for are +1 and -6.

In conclusion, $$n^8-5n^4-6=(n^4+1)(n^4-6)$$.