Factor Completely

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Gr8fu13

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Feb 13, 2011
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The instructions are to factor completely (type your answer in factored form):
4x^8 - 24x^7 + 16x^6

Would I use the method of letting U=x^6
Or since 4 goes into all of these numbers, rewrite like:
4x^6(x^2 - 6x + 4)

I tried using the quadratic formula but I ended up with -13 +- Square root of 320 all over 80. I don't know where to go from there. Can someone please assist me? I cannot grasp this concept AT ALL. Something ,must be wrong with my BRAIN!
 
Gr8fu13 said:
The instructions are to factor completely (type your answer in factored form):
4x^8 - 24x^7 + 16x^6

Would I use the method of letting U=x^6.....NO....
Or since 4 goes into all of these numbers, rewrite like:
4x^6(x^2 - 6x + 4).....YES!!! 4x[sup:10u8gbhu]6[/sup:10u8gbhu] is a factor that is common to all of the terms of the original trinomial. In MY humble opinion, removing the greatest common factor of all of the terms should always be the first factoring step you take.

I tried using the quadratic formula but I ended up with -13 +- Square root of 320 all over 80. I don't know where to go from there. Can someone please assist me? I cannot grasp this concept AT ALL. Something ,must be wrong with my BRAIN!
Click to expand...

Ok...now you have 4x[sup:10u8gbhu]6[/sup:10u8gbhu] * (x[sup:10u8gbhu]2[/sup:10u8gbhu] - 6x + 4)

Your next step should be to work on the trinomial, x[sup:10u8gbhu]2[/sup:10u8gbhu] - 6x + 4, to see if it can be factored further. If it can't be, you're DONE.
 
Nothing is wrong with your brain except as tkhunny says you sabotage yourself by saying you can't all the time.

Now TWO issues, one mechanical and one conceptual.

Mechanical: how to use the quadratic formula.

You factored (4x^8) - (24x^7) + (16x^6) = (4x^6)(x^2 - 6x + 4). That was an EXCELLENT first step; you did well.

Now WHAT part of the expression on the right is a quadratic? You can only apply the quadratic formula to quadratics, right?

That is step 2. You CAN do it. Please tell me your answer.
 
I am not sure exactly what your asking. The equation on the right is x^2 - 6x + 4. I know that a=1, b=-6, c=4. The quadratic equation has to be in the form of ax^2 + bx + c in order to use the quadratic formula.
 
Gr8fu13 said:
I am not sure exactly what your asking. The equation on the right is x^2 - 6x + 4. I know that a=1, b=-6, c=4. The quadratic equation has to be in the form of ax^2 + bx + c in order to use the quadratic formula.
Click to expand...

You answered my question by IDENTIFYING THE QUADRATIC EXPRESSION (not equation). You cannot apply the quadratic formula to anything except a quadratic obviously. So you have to recognize that you have a quadratic before the quadratic formula will pop into your head.

Next step: what is the discriminant in the quadratic formula?
 
And the discriminant of the expression you actually have to deal with this for THIS problem is?
 
YES YES YES

Now here is a trick to learn. The first thing to do with the quadratic formula is to look at the discriminant. If it is negative, there are no real real roots to the quadratic. You know this I am sure.

If the discriminant is not a perfect square, there are no rational roots to the quadratic. (You know the difference between rational and real numbers I suppose.)

20 is not a perfect square. When you take its square root, you get 2 * the square root of 5. So it IS possible to factor your expression, but not into rational numbers, which is what I suspect your teacher wants.

So you have solved THIS problem. The fact that the discriminant is NOT a non-negative perfect square means that you CANNOT factor the quadratic using rational numbers.

So YOU know how to solve this problem, and there is nothing at all wrong with your brain.

UH OH: FACTOR IT COMPLETELY. Maybe your teacher DOES want you to factor it using real numbers, rather than just rational numbers. Is that possible?

What you probably do not know is why the quadratic formula is relevant. If you have the time I will try to explain why it is relevant. Learning formulas without learning reasons is a horrible way to learn math, but you have other subjects and friends and stuff so if you do not have time right now, just tell me.
 
I don't think it can be factored any further. Can it? I cannot think of anything that would have a product of 4 and a sum of -6. If I tried to factor further I would end up with 6 + square root of 20 over 2, and 6 - square root of 20 over 2. Am I completely off here??
 
Gr8fu13 said:
I don't think it can be factored any further. Can it? I cannot think of anything that would have a product of 4 and a sum of -6. If I tried to factor further I would end up with 6 + square root of 20 over 2, and 6 - square root of 20 over 2. Am I completely off here??
Click to expand...

Yes, you are. It was what, in my fumbling way, I was trying to explain in my previous post. If the discriminant is not the square of a rational number, there is no factorization using rational numbers. If, however, the dicriminant is non-negative, there is a factorization using the real numbers

[x + 3 + sqr(5)] * [x + 3 - sqr(5)] = x^2 + x[3 - sqr(5)] + x[3 + sqr(5)] + [3 + sqr(5)][3 - sqr(5)]
x^2 + 3x - xsqr(5) + 3x + xsqr(5) + 9 - 5 = x^2 + 6x + 4.

PS As I said, little of this is going to make sense until you know WHY the quadratic formula is relevant to factoring a quadratic. Let me know when you have time for that
 
So you end up with:

4x^6(x - 3 + r)(x - 3 - r) where r = sqrt(5)
 
Gr8fu13 said:
b^2 - 4 (a) (b) = discriminent
Click to expand...

No, Gr8fu13, the discriminant = \(\displaystyle b^2 - 4ac.\)


Gr8fu13 said:
-6^2 - 4 (1) (4) = 36 - 16 = 20
Is this right?
Click to expand...

JeffM said:
YES YES YES
Click to expand...

No, Gr8fu13 and JeffM, it is not correct.

Applying the formula for the discriminant gives:

(-6)^2 - 4(1)(4) = . . . . . . You are substituting -6 in for b, so it must have grouping symbols around it.

36 - 16 =

20


\(\displaystyle -6^2 \ne (-6)^2 . \ . \ . \ . \ . \text{This \ is \ one \ of \ the \ more \ common \ errors \ in \ algebra.}\)

\(\displaystyle -6^2 \ = \ -(6^2) \ = \ -(36) \ = \ -36\)

\(\displaystyle (-6)^2 \ = \ (-6)(-6) \ = \ 36\)
 
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