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, \(n^8-5n^4-6=(n^4+1)(n^4-6)\).