Practice Factoring General Polynomials

Factor completely.

1. $$y^{10}+7y^5-8=$$

2. $$-5h^4-15h^3-10h^2=$$

3. $$x^2-3xy-10y^2=$$

4. $$n^8-5n^4-6=$$

Source: Khan Academy, https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-factoring-polynomials-quadratic-forms/e/factoring_polynomials_2
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1. 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, $$y^{10}+7y^5-8=(y^5+8)(y^5-1)$$.

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

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

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