SOLUTION: Help! I feel like i'm up to my eyeballs in polynomials! I got a test and I need help with these 4 problems. #1 and #2 need to be factored. #3 and #4 need to be solved. Thanx =0) 1)

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Question 39580: Help! I feel like i'm up to my eyeballs in polynomials! I got a test and I need help with these 4 problems. #1 and #2 need to be factored. #3 and #4 need to be solved. Thanx =0) 1) 8x^3+8x-6x
2) 24x^2-92x+80
3) 6x^2+13x+6=0
4) 2(a^2+9)=15a
Found 2 solutions by tran3209, stanbon:
Answer by tran3209(100) (Show Source):
You can put this solution on YOUR website!
1-) 8x^3+8x-6x --->8x^3+2x --->2x(4x^2+1)
Anwser 2x(4x^2+1)
Note: I think you might have copied the problem down wrong. Instead of 8x
it is 8x^2.
2-)24x^2-92x+80 --->4(6x^2-23x+20)--->4(3x-4)(2x-5)
Anwser: 4(3x-4)(2x-5)
3-)6x^2+13x+6=0 --->(2x+3)(3x+2)=0
2x+3=0 ====> 2x=-3 ==>x=-3/2
3x+2=0 ====> 3x=-2 ==>x=-2/3
Anwser x=-3/2 and x=-2/3
4-)2(a^2+9)=15a ---> 2a^2+18=15a --->2a^2-15a+18=0
(2a-3)(a-6)=0
2a-3=0 --->2a=3 --->a=3/2
a-6=0 --->a=6
Anwser a=3/2 and a=6

Answer by stanbon(48579) (Show Source):
You can put this solution on YOUR website!
1) 8x^3+8x^2-6x
=2x(4x^2+4x-3)
=2x(2x-1)(2x+3)
2) 24x^2-92x+80
4(6x^2-23x+20)
=4(?)
Use quadratic formula to get:
x=[23+-sqrt(1009]/12
=4(x-(23+sqrt(1009)]/12)(x-(23)-sqrt(1009))/12)
3) 6x^2+13x+6=0
x=[-13+-sqrt25]/12
4) 2(a^2+9)=15a
2a^2+18-15a=0
2a^2-15a+18=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=81 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6, 1.5. Here's your graph:

Hope this helps.
Cheers,
Stan H.