SOLUTION: Factor completely:
x^3 – 2x^2 + 6x – 12
. Factor completely:
x^2 + 17x + 16
Factor completely:
8x^2 + 6x – 5
Algebra.Com
Question 131837: Factor completely:
x^3 – 2x^2 + 6x – 12
. Factor completely:
x^2 + 17x + 16
Factor completely:
8x^2 + 6x – 5
Found 2 solutions by stanbon, edjones:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Factor completely:
x^3 – 2x2 + 6x – 12
= x^2(x-2) + 6(x-2)
=(x-2)(x^2+6)
------------------
. Factor completely:
x^2 + 17x + 16
= (x+16)(x+1)
-------------------
Factor completely:
8x^2 + 6x – 5
= 8x^2+10x-4x-5
= 2x(4x+5)-(4x+5)
= 4x+5)(2x-1)
=================
Cheers,
Stan H.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
x^3 – 2x^2 + 6x – 12
x^2(x-2)+6(x-2) Factoring by grouping.
(x^2+6)(x-2)
.
x^2 + 17x + 16
(x+16)(x+1)
.
8x^2 + 6x – 5
8*-5=-40
What 2 factors -40 when added equal +6? Ans: 10, -4
8x^2+10x-4x-5 Put -4x and 10xin the middle in either order.
2x(4x+5)-(4x-5) Factoring by grouping again.
(2x-1)(4x+5)
.
Ed
RELATED QUESTIONS
Factor completely:... (answered by mananth)
3. Factor completely:
x^2 + 17x + 16
(answered by edjones)
Completely factor 2x^3 + 6x^2 -... (answered by fcabanski)
factor completely: 2x^2-12x+18
factor completely: 2x^2-9
factor completely:... (answered by stanbon)
Factor completely: x^3 - x^2 -... (answered by Earlsdon)
5. Factor completely:
8x^2 + 6x – 5
(answered by edjones)
Factor Completely:... (answered by edjones)
Factor completely. Show all work necessary.
6x^2 + 17x... (answered by richwmiller)
Factor completely: x^2 - 8x +... (answered by jim_thompson5910)