trinomials using Ac_method

[Lisac_101]

I need help As Soon As Possible please! I have to use the AC-method to factor out 15x⁴-16x²+4, and im not sure where to start, because i know you are suppose to start will combining like terms, and spliting the middle term into two parts. I am confused of how to do this being that i cant find anything common, and not sure how to slpit the middle term. i tried with multiplying to two end numbers, then finding two numbers that add to it, but the answer did not come out right.

 

[mmm4444bot]

i tried with multiplying to two end numbers, then finding two numbers that add to it, but the answer did not come out right.

Did you double-check your arithmetic?

Please show us the beginnings of what you tried.

Cheers :cool:

 

[soroban]

Hello, Lisac_101!

I think you have the "AC method" confused . . .

\(\displaystyle \text{Factor }15x^4 -16x^2+4\)

We have the trinomial: .\(\displaystyle Ax^2 + Bx + C\)

Multiply \(\displaystyle A\) by \(\displaystyle C\).

Factor \(\displaystyle AC\) into two parts whose sum is the middle coefficient.
That is, factor \(\displaystyle 60\) into two parts whose sum is \(\displaystyle 16.\)

Here we go . . .

. . \(\displaystyle \begin{array}{ccc}\text{Factors} & \text{Sum} \\ \hline 1\cdot60 & 61 & \text{no} \\ 2\cdot30 & 32 & \text{no} \\ 3\cdot20 & 23 & \text{no} \\ 4\cdot15 & 19 & \text{no} \\ 5\cdot12 & 17 & \text{no} \\ 6\cdot10 & 16 & \leftarrow\text{ yes!} \end{array}\)


The middle term is \(\displaystyle -16x^2\), so we will use \(\displaystyle -6x^2\) and \(\displaystyle -10x^2.\)

\(\displaystyle \begin{array}{cc}\text{Replace the middle term:} & 15x^4 - 6x^2 - 10x^2 + 4 \\ \text{Factor by grouping:} & 3x^2(5x^2-2) - 2(5x^2-2) \\ \text{Factor out common factor:} &(5x^2-2)(3x^2-2) \end{array}\)

 

[Lisac_101]

I figured it out, but thanks anyways.

 

[lookagain]

The middle term is \(\displaystyle -16x\), so we will use \(\displaystyle -6x\) and \(\displaystyle -10x.\)

\(\displaystyle > > 15x^4 - 6x^2 - 10x^2 + 4 < < \)

I removed my post after mmm4444bot's, because it was much the same as soroban's.
I didn't want to be guilty of giving away the solution.


Anyway, it doesn't have to be rewritten as above.

It can also be rewritten as \(\displaystyle 15x^4 - 10x^2 - 6x^2 + 4\)
before factoring by grouping.



This was edited because the exponents were wrong.

 

[Deleted member 4993]

I removed my post after mmm4444bot's, because it was much the same as soroban's.
I didn't want to be guilty of giving away the solution.


Anyway, it doesn't have to be rewritten as above.

It can also be rewritten as \(\displaystyle 15x^2 - 10x - 6x + 4\)
before factoring by grouping.

The problem posted by OP was:

15x⁴ - 16x² + 4

I think the proper - less confusing way to do this would be to substitute

u = x²

Then factorize, to get,

15u² - 16u + 4 → 15u² - 10u - 6u + 4 → (5u - 2)(3u - 2) → (5x² - 2)(3x² -2)

Further factorization can be conducted as needed.