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.