# How can i factor f(x) = 2x^2 + x - 6

• 01-29-2015, 10:29 PM
abel muroi
How can i factor f(x) = 2x^2 + x - 6
I'm trying to convert the function into the standard from but i cant even factor this correctly

f(x) = 2x2 + x - 6

at first i thought 2(x2 + x) - 6 would equal 2x2 + x - 6 but then i miraculously found out that the product of 2 and x is NOT x.

can anyone tell me how to factor this?
• 01-29-2015, 10:42 PM
Jomo
Quote:

Originally Posted by abel muroi
I'm trying to convert the function into the standard from but i cant even factor this correctly

f(x) = 2x2 + x - 6

at first i thought 2(x2 + x) - 6 would equal 2x2 + x - 6 but then i miraculously found out that the product of 2 and x is NOT x.

can anyone tell me how to factor this?

Even if 2x2 + x - 6 equaled 2(x2 + x) - 6 the rhs is not factored! Something is factored if things (not inside parenthesis) are multiplied together. You are subtracting 6.

I am assuming that you want to do what you did with the last problem? Please, in the future tell us what you want.

In any case you need to understand that to undo multiplying by 2 you simply divide by 2.

2x^2 + x -6 = (2x^2 + x ) -6 = 2(x^2 + x/2) -6. Now that x/2 =1x/2 =(1/2)x. Now complete the square and then show us your work.
• 01-29-2015, 10:47 PM
abel muroi
Quote:

Originally Posted by Jomo
Even if 2x2 + x - 6 equaled 2(x2 + x) - 6 the rhs is not factored! Something is factored if things (not inside parenthesis) are multiplied together. You are subtracting 6.

I am assuming that you want to do what you did with the last problem? Please, in the future tell us what you want.

In any case you need to understand that to undo multiplying by 2 you simply divide by 2.

2x^2 + x -6 = (2x^2 + x ) -6 = 2(x^2 + x/2) -6. Now that x/2 =1x/2 =(1/2)x. Now complete the square and then show us your work.

Yes in the future i will remember to be more specific. Thank you

and here is my work..

f(x) = 2x2 + x - 6
f(x) = 2(x2 + 1/2x) - 6
f(x) = 2(x2 + 1/2x + 1/16 - 1/6) - 6 ((i divided 1/2 by 2 and then i squared it which got me 1/16, then i added and subtracted 1/16))
f(x) = 2(x2 +1/2x + 1/16) -1/8 - 6 (( i multiplied 2 and -1/6 which got me -1/8))
f(x) = 2(x + 1/4)2 - 49/8

so the vertex is (-1/4, -49/8)

or (-1/4 , -6 1/8)
• 01-29-2015, 11:46 PM
ClaireWu
Quote:

Originally Posted by abel muroi
I'm trying to convert the function into the standard from but i cant even factor this correctly

f(x) = 2x2 + x - 6

at first i thought 2(x2 + x) - 6 would equal 2x2 + x - 6 but then i miraculously found out that the product of 2 and x is NOT x.

can anyone tell me how to factor this?

2x^2 + x - 6 = 2x^2 + 4x - 3x - 6

= 2x(x + 2) - 3(x + 2) = (x +2)(2x - 3) (Ans)

Or by completing the square method,

2x^2 + x - 6 = 2[x^2 + (1/2)x] - 6

=2[x^2 + (1/2)x + (1/4)^2 - (1/4)^2] - 6

=2[(x+1/4)^2 - 1/16] - 6

=2(x + 1/4)^2 - 1/8 - 6

=2(x + 1/4)^2 - 49/8

=2[(x + 1/4)^2 - 49/16]

=2[(x + 1/4)^2 - (7/4)^2]

=2[x + 1/4 + 7/4][x + 1/4 -7/4]

=2(x + 2)(x - 3/2)

=(x + 2)(2x - 3) (Ans)
• 01-29-2015, 11:52 PM
Jomo
Quote:

Originally Posted by ClaireWu
2x^2 + x - 6 = 2x^2 + 4x - 3x - 6

= 2x(x + 2) - 3(x + 2) = (x +2)(2x - 3) (Ans)

Very nice! The thing is that the OP used the wrong terminology as he does not want to factor f(x) but rather he wants to write f(x) in the form A(x-B)^2 +C
• 01-30-2015, 12:02 AM
ClaireWu
Quote:

Originally Posted by Jomo
Very nice! The thing is that the OP used the wrong terminology as he does not want to factor f(x) but rather he wants to write f(x) in the form A(x-B)^2 +C

ok
• 01-30-2015, 12:53 AM
Ishuda
Quote:

Originally Posted by abel muroi
I'm trying to convert the function into the standard from but i cant even factor this correctly

f(x) = 2x2 + x - 6

at first i thought 2(x2 + x) - 6 would equal 2x2 + x - 6 but then i miraculously found out that the product of 2 and x is NOT x.

can anyone tell me how to factor this?

ClaireWu provided a method. Another is to just use the quadratic formula if you can't 'see the roots fooling around'. That is, if x1 and x2 are roots to
f(x) = a x2 + b x + c
then
f(x) = a (x - x1) (x - x2)

For your problem we have the roots given by
$$\frac{-1 \pm \sqrt{1^2 - 4 (2) (-6)}}{2 (2)}= \frac{-1 \pm \sqrt{49}}{4} = \frac{-1 \pm 7}{4}$$
which gives -2 and 3/2 as the roots
So
f(x) = 2x2 + x - 6 = 2 (x + 2) (x - 3/2) = (x + 2) (2 x - 3)

Which, thankfully, is the same thing given above.
• 01-30-2015, 09:45 AM
HallsofIvy
Quote:

Originally Posted by abel muroi
I'm trying to convert the function into the standard from but i cant even factor this correctly

f(x) = 2x2 + x - 6

at first i thought 2(x2 + x) - 6 would equal 2x2 + x - 6 but then i miraculously found out that the product of 2 and x is NOT x.

can anyone tell me how to factor this?

As far as factoring is concerned (not "standard form" or "complete the square"), I would note that 2x2 can be factored as (2x)(x) and -6 as either (-3)(2) or (3)(-2).

So possible factors are (2x+ 3)(x- 2), (x- 3)(2x+ 2), (x+ 3)(2x- 2), and (2x- 3)(x+ 2).

Check them:
(2x+ 3)(x- 2)= (2x)(x)+ (2x)(-2)+ (3)(x)+ (3)(-2)= 2x2- 4x+ 3x- 6= 2x2- x- 6. No, that is not correct. It has "-x" rather than "+x".

(x- 3)(2x+ 2)= (x)(2x)+ (x)(2)+ (-3)(2x)+ (-3)(2)= 2x2+ 2x- 6x- 6= 2x2- 4x- 6. No, that is not correct. It has "-4x" rather than "+x".

(x+ 3)(2x- 2)= (x)(2x)+ (x)(-2)+ (3)(2x)+ (3)(-2)= 2x2- 2x+ 6x- 6= 2x2+ 4x- 6. No, that is not correct. It has "+4x" rather than "+x".

(2x- 3)(x+ 2)= (2x)(x)+ (2x)(2)+ (-3)(x)+ (-3)(2)= 2x2+ 4x -3x- 6= 2x2+ x- 6. YES, that is correct!
• 01-30-2015, 12:51 PM
Ishuda
Quote:

Originally Posted by Jomo
Very nice! The thing is that the OP used the wrong terminology as he does not want to factor f(x) but rather he wants to write f(x) in the form A(x-B)^2 +C

But that was provided by ClaireWu at the point
Quote:

...
=2(x + 1/4)^2 - 49/8
...