SOLUTION: Factor completely. 1. 6x^6 + 8x^4 – 15x^3 – 20x 2. 9x^2 – 27x + 20 3. 2x^2 – 128

Algebra.Com
Question 401301: Factor completely.
1. 6x^6 + 8x^4 – 15x^3 – 20x


2. 9x^2 – 27x + 20

3. 2x^2 – 128

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
Factor completely.
1. 6x^6 + 8x^4 – 15x^3 – 20x

2. 9x^2 – 27x + 20
3. 2x^2 – 128
1. 6x^6 + 8x^4 – 15x^3 – 20x
=2x^4(3x^2+4)-5x(3X^2+4)
=(3x^2+4)(2x^4-5x)
=(3x^2+4)(2x^3-5)(x)
2. 9x^2 – 27x + 20 (set to zero and solve quadratic equation)
=(3x=5)(3x-4)
3. 2x^2 – 128
=x^2=128/2
=x^2-64 (difference of squares)
=(x+8)(x-8)


RELATED QUESTIONS

factor completely.... (answered by stanbon)
factor... (answered by mananth)
Completely factor 2x^3 + 6x^2 -... (answered by fcabanski)
factor completely... (answered by mananth)
Factor completely. x^3 + 8x^2 + 15x (answered by checkley71)
Factor completely using integer coefficients: e) x^4+8x^3-2x^2-16x f)... (answered by Earlsdon)
How to factor 64x^6 - 1 completely? 4/3(6x-3) = 3/4(8x-12) Solve for x: -2/9x... (answered by josgarithmetic)
Factor by grouping: 6x^3-8x^2-15x+20 (answered by Alan3354)
Factor completely Show your work.... (answered by checkley77)