Question 1097337: Factor
6(p-5)^2+11(p-5)+3
I KNOW A*C= 18
AND THE SUM IS 11
NEED HELP WITH THE STEPS AFTER THAT PLEASE
Found 2 solutions by Boreal, KMST:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! let u=p-5
This is now 6u^2+11u+3
=(3u+1)(2u+3),
which is (3(p-5)+1)(2(p-5)+3)
(3p-14)(2p-7)
Check
The last is 6p^2-49p+98
The original is 6p^2-60p+150+11p-52
That is 6p^2-49p+98
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! You need a pair of factors of 18 that add up to 11.
The pairs of factors for 18 are ;
1 and 18 (adding to 19),
2 and 9 (adding to 11),
3 and 6 (adding to 9).
So, your pair of factors is 2 and 9.
Next, you could re-write the polynomial uswing 2 and 9 as coefficients for a split middle term as
6(p-5)^2+2(p-5)+9(p-5)+3 ,
and then factor by parts
6(p-5)^2+11(p-5)+3 = 6(p-5)^2+2(p-5)+9(p-5)+3 = [6(p-5)^2+2(p-5)]+[9(p-5)+3]
= 2(p-5)[3(p-5)+1] + 3[3(p-5)+1] = [2(p-5)+3][3(p-5)+1]
= (2p-10+3)(3p-15+1) = (2p-7)(3p-14)
It looks complicated, and looks like it would be easy to make some mistake.
An easier option would be to start by making a cahnge of variable,
with y = p-5 , and factor 6y^2+11y+3.
6y^2+11y+3 = 6y^2+2y+9y+3 = (6y^2+2y) + (9y+3) = 2y(3y+1) + 3(3y+1)
= (2y+3)(3y+1)
At that point, you can change back to p as a variable, rplacing p-5 for y:
[2(p-5)+3] [3(p-5)+1] = (2p-10+3)(3p-15+1) = (2p-7)(3p-14)
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