Question

8-10x-3x^2/x^2+9x+20 PLEASE SOLVE AND EXPLAIN

Answer 1

have covered factoring of quadratics yet?

Answer 2

yep

Answer 3

.... ok.. one sec

Answer 4

ok

Answer 5

\(\bf \cfrac{8-10x-3x^2}{x^2+9x+20}\implies \cfrac{8-10x-3x^2}{(x+4)(x+5)}\)

see how the denominator was factored?

Answer 6

yes i do understand that. but how do i factor the numerator

Answer 7

\(\large \begin{array}{cccllll}
x^2&+9x&+20\\
&\uparrow &\uparrow \\
&4+5&4\cdot 5
\end{array}\)

the sum of the components, add up to 9
their product gives 20

Answer 8

but like the numerator is backward...how do i get the x^2 on the left side

Answer 9

teh numerator is mainly trial and error

you'd want to factor out the "8", the constant
and factor out the "3", the leading term coefficient

then fiddle and check which pair gives the middle and the last term


so \(\large {
\cfrac{8-10x-3x^2}{x^2+9x+20}\implies \cfrac{8-10x-3x^2}{(x+4)(x+5)}
\\ \quad \\
\cfrac{-(3x^2+10x-8)}{(x+4)(x+5)}\quad \\ \quad \\
\begin{cases}
8=2\cdot 2\cdot \cdot 2\cdot 1\\
3=3\cdot 1
\\\hline\\
(3\cdot +4)+(1\cdot -2)\implies +12-2\implies &10\\
-2\cdot +4\implies &-8\\
3x\cdot x\implies &3x^2
\end{cases}\qquad thus
\\ \quad \\
\cfrac{-(3x^2+10x-8)}{(x+4)(x+5)}\implies \cfrac{-[(3x-2)\cancel{(x+4)}]}{\cancel{(x+4)}(x+5)}
}\)

Answer 10

ok thank you

Answer 11

yw