Question

Can someone show me an easy way to factor
20-48x-36x^2

Answer 1

\[20-48x-36x^2\]

Answer 2

what factors do 20, 48, and 36 have in common?

Answer 3

4?

Answer 4

yes, so we can factor out 4. I'm going to factor out -4 so that x^2 term will turn positive

\[\Large 20-48x-36x^2 = -4(-5+12x+9x^2)\]
\[\Large 20-48x-36x^2 = -4(9x^2+12x-5)\]

hopefully you're seeing how I'm getting this?

Answer 5

\[=-4\left( 9x^2+12x-5 \right)\]
9*-5=-45
15*-3=-45
15-3=12
write 12=(15-3)

Answer 6

yes i see how you got that. how does the answer become (6x+4)^2 ?

Answer 7

thats what i have it equivalent to I'm not sure how to get there

Answer 8

I don't know how you got that

Answer 9

20-48x-36x^2 is not equivalent to (6x+4)^2

Answer 10

ok thank you !!

Answer 11

to factor 9x^2 + 12x - 5, you need to use the AC method or you can use the quadratic formula

AC method:
step 1) multiply the first and last coefficients
9*(-5) = -45

step 2) find two numbers that multiply to -45 and add to 12
these two numbers happen to be +15 and -3

step 3) break up the middle term 12x into +15x and -3x and then factor by grouping

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quadratic formula:
step 1) solve 9x^2 + 12x - 5 = 0 for x using the quadratic formula

step 2) use the roots to construct the factorization

Answer 12

\[-\left\{ 36x^2+48 *4x-20 \right\}=-\left\{ \left( 6x \right)^2+2*6x*4+\left( 4 \right)^2-\left( 4 \right)^2-20 \right\}\]
\[=-\left( 6x+4 \right)^2+36\]

Answer 13

thanks guys! @surjithayer -- are you using completing the square?

Answer 14

for the other method i gave you the hint,and jim_thomson 5910 gave more detailed.
i have done both ways
in the second method
\[a^2-b^2=\left( a+b \right)\left( a-b \right)\]

Answer 15

\[6^2-\left( 6x+4 \right)^2=\left\{ 6+6x+4 \right\}\left\{ 6-\left( 6x+4 \right) \right\}\]

Answer 16

thanks I'm trying the AC method now but not sure how to write it out. this is just a part of an integral problem and the denominator of the integral changes to (6x+4)^2+36 but I'm having a hard time following

Answer 17

\[-4(9x^2+12x-5)\]\[=-4(x^2+12x-45)\]\[=-4\left[ (x+15)(x-3)\right]\]use the leading coefficient, 9, you used to factor out the simplified quadratic back into your equation by dividing both terms by it. \[=-4\left[\left(x+\frac{15}{9}\right)\left(x-\frac{3}{9}\right)\right]\]Simplify your fractions. \[=-4\left[\left(x+\frac{5}{3}\right)\left(x-\frac{1}{3}\right)\right]\]This becomes: \[=-4\left[\left(\frac{3x+5}{3}\right)\left(\frac{3x-1}{3}\right)\right]\]Multiplying 3 to both sides, we can get rid of that common denominator of \(3\). This leaves us with: \[=\boxed{-4(3x+5)(3x-1)}\]