SOLUTION: Factor completely: x^3 – 2x^2 + 6x – 12 . Factor completely: x^2 + 17x + 16 Factor completely: 8x^2 + 6x – 5

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Factor completely: x^3 – 2x^2 + 6x – 12 . Factor completely: x^2 + 17x + 16 Factor completely: 8x^2 + 6x – 5       Log On


   

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) About Me  (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)
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. 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) About Me  (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