factoring trinomials

[SavannahAshley]

factor the trinomials completely
8x^2-14xy+3y^2
(8x-21y)(x-2y)? I think it is wrong? maybe, (8x-8y)(x-3y)
or even (8x-4y)(x-6y)
... or (4x-6y)(2x-4y) or (4x-3y)(2x-8y)



I also have two others
9y^2+48y+64
(9y+6)(y+96)
or (9y+32)(y+18)
9x64,16x36,12x48


and
27x^2+2x-1/9
I hate factions, and I dont even know where to start

 

[Deleted member 4993]

factor the trinomials completely
8x^2-14xy+3y^2
(8x-21y)(x-2y)? I think it is wrong? maybe, (8x-8y)(x-3y)
or even (8x-4y)(x-6y)
... or (4x-6y)(2x-4y) or (4x-3y)(2x-8y)

Multiply your possible answers out and check

I also have two others
9y^2+48y+64
(9y+6)(y+96)
or (9y+32)(y+18)
9x64,16x36,12x48


and
27x^2+2x-1/9
I hate factions, and I dont even know where to start

8x^2-14xy+3y^2

= x^2[8 - 14(y/x) + 3 (y/x)[sup:1db8xa9q]2[/sup:1db8xa9q]]

substitute

y/x = u

= x^2[8 - 14(u) + 3 (u)[sup:1db8xa9q]2[/sup:1db8xa9q]]

=x^2(3u[sup:1db8xa9q]2[/sup:1db8xa9q] - 12u -2u + 8)

= x^2(3u - 2)(u - 4)

= x^2 (3y/x - 2)(y/x - 4)

= (3y - 2x)(y - 4x)

If you hate fractions - why are you torturing yourself with these?

 

[SavannahAshley]

so, I need to take out x^2 in the very beginning and drop it at the very end?
Im just really stuck on this homework :shock:

 

[Denis]

This is factoring.
The work you show indicates you don't know the basics.
Did you miss the math classes on factoring?

 

[stapel]

factor the trinomials completely

...I dont even know where to start

To learn how to factor, start here, then read this, and finish up with the formulas listed here. :wink:

 

[SavannahAshley]

Thank you soo much Stapel!

This is factoring.
The work you show indicates you don't know the basics.
Did you miss the math classes on factoring?

I know how to factor, Im lost when it comes to putting a number as a in the form ax^2+bx+c
I was on tutorvista.com looking for help and they said when factoring, with a larger number than 1 in the a position, that you had to multiply it by c and find the factors using that product.
My teacher today, said they just confused me.

 

[Mrspi]

Thank you soo much Stapel!

This is factoring.
The work you show indicates you don't know the basics.
Did you miss the math classes on factoring?

I know how to factor, Im lost when it comes to putting a number as a in the form ax^2+bx+c
I was on tutorvista.com looking for help and they said when factoring, with a larger number than 1 in the a position, that you had to multiply it by c and find the factors using that product.
My teacher today, said they just confused me.

That's the method I use and teach to my students.

Suppose you have to factor this trinomial:

6x[sup:1o1yfls8]2[/sup:1o1yfls8] + x - 12

Multiply the coefficient of the x[sup:1o1yfls8]2[/sup:1o1yfls8] by the constant term (that's the "a times c" you mentioned.) The coefficient of x[sup:1o1yfls8]2[/sup:1o1yfls8] is 6, and the constant term is -12. 6*(-12) = -72

NOW...look for two numbers which multiply to -72, and which add up to the "middle" coefficient, which is 1 (because you have x, or 1x, as the middle term).

-8*9 = -72, and (-8) + 9 = 1....so the numbers we want are -8 and 9. Use these numbers to rewrite the middle term as -8x + 9x:

6x[sup:1o1yfls8]2[/sup:1o1yfls8] - 8x + 9x - 12

Now, factor by grouping the first two terms together, and the last two terms together. The first two terms have a common factor of 2x. The last two terms have a common factor of 3. Remove the common factor from each pair of terms:

2x*(3x - 4) + 3(3x - 4)

Now, you have two terms, and they have a common factor of (3x - 4). Remove that factor, and you have

(3x - 4)(2x + 3)

That's the factorization you're looking for, and you can verify that it is correct by multiplying the two factors together.

If your teacher has a "better idea," then your teacher is the person from whom you need to seek help.