simplify n^2+8n+16/n^2-6n+9 divided by n+4/n-3

Question

pfenn1 · Answer

Is this the correct expression?$$\frac{n^2+8n+16}{n^2-6n+9}\div \frac{n+4}{n-3}$$

anonymous · Answer

yes.

pfenn1 · Answer

Okay that expression is equivalent to $$\frac{n^2+8n+16}{n^2-6n+9}\ \times\frac{n-3}{n+4}$$

pfenn1 · Answer

Can you factor the expression$$n^2+8n+16=(n+a)(n+b)$$

anonymous · Answer

i totally dont get how to do any of this. sorry.

pfenn1 · Answer

we know that often a binomial expression like $$n^2+8n+16=(n+a)(n+b)=n^2 +(a+b)n +ab$$so we have to find a and b such that a+b = 8 and ab = 16.

pfenn1 · Answer

Let's try a=4 b=4. $$n^2+8n+16=(n+4)(n+4)$$That works. Now we have to factor the denominator$$n^2-6n+9=(n+c)(n+d) $$where c+d = -6 and cd = 9.  Can you do this one?

anonymous · Answer

So what is the answer?

pfenn1 · Answer

Well, c = -3 and d = -3 works.  so $$n^2-6n+9=(n-3)(n-3)=(n-3)^2$$ Now we can rewrite the original expression from above as$$\frac{n^2+8n+16}{n^2-6n+9}\ \times\frac{n-3}{n+4}=\frac{(n+4)^2}{(n-3)^2} \times \frac{n-3}{n+4}$$which can be simplified to$$\frac{n+4}{n-3}$$

anonymous · Answer

thank you:)