need some factoring help

[billgrenz]

Hi I have a page of factoring and I got most of the problems except for these 4 questions that are stumping me. Can anyone solve these for me?

1. 3y^2 - 2y -5 the ^2 is supposed to mean squared

2. x^2 - 6x + 10

3.x^6 - y^4

4.2x^2 -5x - 12

Thank you for your help!

 

[Deleted member 4993]

Hi I have a page of factoring and I got most of the problems except for these 4 questions that are stumping me. Can anyone solve these for me?

1. 3y^2 - 2y -5 the ^2 is supposed to mean squared

2. x^2 - 6x + 10
Use quadratic equation

Ax[sup:3fdaafol]2[/sup:3fdaafol] + Bx + C = A*(x-x[sub:3fdaafol]1[/sub:3fdaafol])(x-x[sub:3fdaafol]2[/sub:3fdaafol])

where

x[sub:3fdaafol]1,2[/sub:3fdaafol] = [-B ± ?(B[sup:3fdaafol]2[/sup:3fdaafol]-4AC)]/(2A)



3.x^6 - y^4

use

a[sup:3fdaafol]2[/sup:3fdaafol] - b[sup:3fdaafol]2[/sup:3fdaafol] = (a+b)(a-b)

4.2x^2 -5x - 12

Use quadratic equation as in first two problems

Thank you for your help!

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[fasteddie65]

1. 3y^2 - 2y - 5 the ^2 is supposed to mean squared
Multiply 3 by -5: -15
Are there factors of -15 whose sum is -2?
Yes; -5 + 3 = -2
Rewrite 3y^2 - 2y - 5 as 3y^2 - 5y + 3y - 5 = (3y^2 - 5y) + (3y - 5) = y(3y - 5) + 1(3y - 5) = (3y - 5)(y + 1)

2. x^2 - 6x + 10
Are there factors of 10 whose sum is -6?
10 = 5 • 2 = -5 • -2 = 1 • 10 = -1 • -10
None of these factors add to -6.
So this trinomial is prime.

3.x^6 - y^4
This is a difference of squares: (x^3)^2 - (y^2)^2 = (x^2 + y^2)(x^3 - y^2)

4.2x^2 - 5x - 12
Again, are there factors of -24 whose sum is -5? How about -8 and 3?
Now, do a similar method as I showed in #1.