Question

Using complete sentences, explain each step in simplifying the ratio ->

Answer 1

factorize both the numerator and denominator ,
cancel common terms

Answer 2

\[\frac {x^{2}-x-20}{x^{2}+8x+16}\]

Answer 3

can you can factorize
\[x^2-x-20=(x+\cdots)(x-\cdots)\]

Answer 4

Could you help me please?

Answer 5

I don't know if you can.. I don't know how to factorize..

Answer 6

ok so the two numbers you want add to -1, and multiply to give 20

Answer 7

**multiply to give -20

Answer 8

the factors of -20 are \[\{20,10,5,4,2,1,-20,-10,-5,-4,-2,-1\}\]

find two of these that add to -1

Answer 9

Wait..

Answer 10

Explain in detail. You want me to pick 2 of those numbers that add to -1 and when the two numbers are multiplied they equal -20?

Answer 11

5 and -4.

Answer 12

almost

Answer 13

-4 + 5 = 1... hmm no.

Answer 14

4 and -5?

Answer 15

you got it!

Answer 16

Okay before you go on.

Answer 17

How do I know the list to get the numbers from in other problems?

Answer 18

hence
\[\Rightarrow x^2-x-20=(x+4)(x-5)\]
and you can even check
\[(x+4)(x-5)=x(x-5)+4(x-4)\]\[\qquad\qquad\qquad=x^2-5x+4x-20\]\[\qquad\qquad\qquad=x^2-x-20\]

Answer 19

I mean like, the list of numbers you gave me {20,10,5,4,2,1,−20,−10,−5,−4,−2,−1}, on other problems, how would I KNOW which number to factorize? Also to add, I don't know which other number I need to look for in other problems, like -1 in this one.

Answer 20

So the next spet is to factorise the denominator\[{x^{2}+8x+16}=(x+\cdots)(x+\cdots)\]
like in the last example we want the missing numbers to multiply to give the constant term (in this example 16) {in the last example 20}

and the sum of the numbers to be the number in front of the \(x\) term,
(in this example 8, ){in the last example -1 [because \(-x=-1\times x\)]}

Answer 21

so factorize the constant term
16 has factors \(\{,,,,\}\) only 5 factors

Answer 22

.. I'm lost.

Answer 23

\[1\times16=16\]
so one and sixteen are both factors,
can you find the other three numbers?

Answer 24

look on some times tables for \(x\times y= 16\)

\(x\) and \(y\) willl be factors

Answer 25

I don't know man..
I don't get this..

Answer 26

When you have done factorizing both the numerator and denominator\[\frac {x^{2}-x-20}{x^{2}+8x+16}=\frac {(x+4)(x-5)}{(x+\cdots)(x+\cdots)}\]

Some of the factors in the numerator will be the same
as the factors in the denominator.
these common factors can be canceled out of the fraction\[\qquad\qquad= \frac {\cancel{(x+4)}(x-5)}{\cancel{(x+\cdots)}(x+\cdots)}\]\[\qquad\qquad= \frac {(x-5)}{(x+\cdots)}\]..And you have simplified the ratio\[\qquad\qquad= \frac {x-5}{x+\cdots}\]