Question

Solve for x in the equation: 16x^3 + 4x^2 - 4x -1
(all solutions are rational and between +- 1

Answer 1

Factor by grouping, then use the zero product property.

Answer 2

You yet to have an equation where is the = sign?

Answer 3

Good catch, radar, I assumed equal to zero....

Answer 4

I bet you're right @AnimalAin

Answer 5

...and we all know where assumptions get us...

Answer 6

Yes, I did not want to break it down that far lol

Answer 7

If it is equal to zero, one of the roots might be 1/2

Answer 8

In fact it is by inspection.

Answer 9

If it were equal to zero that is.

Answer 10

Continuation of this assumption and dividing the expression by (x-1/2) gives us a quadratic of\[16x ^{2}+12x+2\]

Answer 11

On the summer work assignment that my teacher created it doesn't have an equal sign, yet does say equation,so do I assume or not?

Answer 12

I would think so. In that case,\[16x^3 + 4x^2 - 4x -1=0 \implies 4x^2(4x+1)-(4x+1)=0\]\[\implies 16(x+\frac{1}{2})(x-\frac{1}{2})(x+\frac{1}{4})=0 \implies x \in \left\{ -\frac{1}{2},-\frac{1}{4},\frac{1}{2} \right\}\]

Answer 13

well it won't let me use the editor. 16x^2 +12x +2 =0 factor that and get more roots. Your teacher has put you in a position that you probably have to assume the expression is equal to 0. Chastise her for me when you see her.

Answer 14

From Mathematica:\[16 x^3+4 x^2-4 x-1=(2 x-1) (2 x+1) (4 x+1) \]

Answer 15

Lills, you understand this?

Answer 16

I see where your answer is coming from, but I'm not sure if given another problem I would be able to do it alone

Answer 17

Look for the patterns. In this case, we have a cubic with coefficients that allow factoring by grouping. After that, we have one factor that is a difference of squares. When the whole kebab is factored, you just use the zero product property.

Answer 18

What I am not completely following is how you got from the original equation to the next step, I understand the zero part

Answer 19

Note that the ratio between the first and second coefficients is the same as the ratio between the third and fourth coefficients. This indicates we can factor by grouping. A little more on this here
http://coolmath.com/algebra/04-factoring/07-factor-by-grouping-01.htm

or about halfway down the page here

http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx

Answer 20

Thanks you so much