Solve by factoring 2x^2-2x-12=0?

Question

2x^2-2x-12=0

(2x+6)(x-2)

2x+6=0
2x=6
x=3

x-2=0
x=2

I THINK the two answers for this question are 3 and 2. The thing is, I was taught that the two numbers that multiply to the last number (-12) must also add up to the middle number (2). +6-2=4 and not 2. So how do I know what two proper numbers to use?

Answer 1

If you first factor out a common 2, then you have:
2 (x² - x - 6)

Now this factors to:
2 (x - 3)(x + 2)

So your answers are x = 3 and x = -2

As a double-check:
2(3²) - 2(3) - 12 = 0
2(9) - 6 - 12 = 0
18 - 6 - 12 = 0
18 - 18 = 0 <-- check, so x = 3

2(-2)² - 2(-2) - 12 = 0
2(4) - (-4) - 12 =0
8 + 4 - 12 = 0
12 - 12 = 0 <-- check, so x = -2

Answer 2

First: find a number divisible by the coefficients > (2). Take (2) and bring it to the front; multiply (2) by missing terms in the parenthesis:

2(x^2 - x - 6) = 0

Second: factor the inside. Take the first coefficient and multiply it by the last coefficient = (1)(-6) = -6. Find two numbers that give you (-6) when multiplied and -1 (middle coefficient) when added/subtracted. The numbers are: -3, 2

2(x - 3)(x + 2) = 0

Third: set each term to equal zero > exclude 2 > it doesn't = 0

x - 3 = 0
x - 3 + 3 = 0 + 3
x = 3

x + 2 = 0
x + 2 - 2 = 0 - 2
x = -2

(3, -2)

Answer 3

Solve 2x 12

Answer 4

2x² - 2x - 12 = 0

2x² - 6x + 4x - 12 = 0

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

(2x + 4)(x - 3)

- - - - - - - - - - -

Roots

2x + 4 = 0

2x + 4 - 4 = 0 - 4

2x = - 4

3x/2 = - 4/2

x = - 2. . .<=. .Not used

- - - - -

x - 3 = 0

x - 3 + 3 = 0 + 3

x = 3

The answer is x = 3

Insert the x value into the equation

- - - - - - - - - - -

x² - 2x - 12 = 0

2(3)² - 2(3) - 12 = 0

2(9) - 6 - 12 = 0

18- 6 - 12 = 0

12 - 12 = 0

0 = 0

- - - - - - -s-

Answer 5

The correct factorization is (2x-6)(x+2)
= 2x^2+4x-6x-12
=2x^2-2x-12

In this case because the equation is not monic (the leading coefficient is 2 not 1) then you need two numbers that multiply to
a * c = 2 * -12 = -24
and add up to b = -2
where a, b, and c are the coefficients of the quadratic ax^2+bx+c

Here, you have -6 and 4 satisfy the two conditions above, mult to -24 and add up to -2

You use the -6 and 4 to break up the middle term and solve by grouping as follows:

The middle term -2x becomes -6x + 4x

2x^2 - 6x + 4x -12
Factor out 2x from the first two terms and 4 from the last two terms:

2x ( x-3) + 4 (x-3) = (2x+4) (x-3) = 0

This gives you x=-2 and x=3

You could also have done:

2x^2 + 4x - 6x - 12

2x (x+2) - 6 ( x+2) = (x+2) (2x-6) again giving you x=-2, x=3

Answer 6

1) The first method:
(2x-6)(x+2)=0
2x=6
x=-2
x= 3 or -2

2) The second method:
Factor out a 2:
2(x^2-x-6)=0
(x+2)(x-3)=0
x+2=0
x-3=0
x=-2 or 3

I hope this helps you with your homework!

Answer 7

2x^2 - 2x - 12 = 0
(2x - 6)(x + 2) = 0
2x - 6 = 0 or x + 2 = 0
2x = 6
x = 3
or
x + 2 = 0
x = -2
x = 3 or x = -2
QED

Answer 8

You're close but (2x+6)(x-2) = 2x^2 + 6x - 4x - 12 = 2x^2 PLUS 2x - 12

You need to reverse the signs: (2x-6)(x+2)

2x-6=0 yields x=3
x+2 = 0 yields x= negative 2

Answer 9

2x²-2x-12=0
Factoring:
x² - x - 6 = 0
(x - 3) (x + 2) = 0
x' = 3
x" = -2
or
Justify:
delta = b² - 4ac
delta = -1² - 4.1.-6
delta = 1 + 24 = 25

x = (-b +/-\/delta) : 2a
x = (-(-1) +/-5) : 2.1
x' = (1 + 5): 2 = 3
x" = (1 - 5) : 2 = -2
Solution: {x E R | x = 3 or x = -2}
Happy New Year!
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Answer 10

You left out a negative sign.

2x+6=0
2x= -6
x= -3