First thing I'd do is check if it CAN be factored, by checking the discriminant b^2 - 4ac, the thing that appears in the quadratic formula under a square root sign. If that's not a perfect square, then you can't factor it.
2x^2 can only be factored one way, as 2x * x. So you know you have
(2x )(x )
When the leading coefficient isn't 1, there's always a little bit of trial and error. Here's how that goes.
15 = 3*5 or 1*15. Let's try 3 and 5 first. The constant term is -15, so the factors have to have opposite signs. It could be -3*5 or 3*(-5). And they could occur in either order. So there are four possibilities using 3 and 5:
(2x + 3)(x - 5)
(2x - 3)(x + 5)
(2x + 5)(x - 3)
(2x - 5)(x + 3)
And basically you have to use FOIL to get the x term of each of those and see if any of them are -7x. As it turns out one of them is, so that is the correct factorization.
If none of those worked, you'd want to try using 1 and 15, for instance (2x + 1)(x - 15).
Comments
Shortcut method
Multiply the number behind the x^2 by the number without an X
2*15=30
find 2 numbers that multiply to 30 but add or subtract to equal the number with the single x
x*y = 30
x+y = -7
10*3 = 30
-10+3 = 7
so plug it in like this
(2x-10)(2x+3)
you can pull out a 2 out of the first one
(x-5)(2x+3) ANSWER
First thing I'd do is check if it CAN be factored, by checking the discriminant b^2 - 4ac, the thing that appears in the quadratic formula under a square root sign. If that's not a perfect square, then you can't factor it.
b^2 - 4ac = 7^2 - 4*2*(-15) = 49 - (-120) = 49 + 120 = 169 = 13^2. OK, that's a square. So it's factorable.
2x^2 can only be factored one way, as 2x * x. So you know you have
(2x )(x )
When the leading coefficient isn't 1, there's always a little bit of trial and error. Here's how that goes.
15 = 3*5 or 1*15. Let's try 3 and 5 first. The constant term is -15, so the factors have to have opposite signs. It could be -3*5 or 3*(-5). And they could occur in either order. So there are four possibilities using 3 and 5:
(2x + 3)(x - 5)
(2x - 3)(x + 5)
(2x + 5)(x - 3)
(2x - 5)(x + 3)
And basically you have to use FOIL to get the x term of each of those and see if any of them are -7x. As it turns out one of them is, so that is the correct factorization.
If none of those worked, you'd want to try using 1 and 15, for instance (2x + 1)(x - 15).
2x² - 7x - 15 = 0
x² - 7/4x = 15/2 + (- 7/4)²
x² - 7/4x = (120 + 49)/16
(x - 7/4)² = 169/16
x - 7/4 = ± 13/4
= x - 7/4 - 13/4, = x - 20/4, = x - 5
= x - 7/4 + 13/4, = x + 6/4, = x + 3/2, = 2x + 3
Answer: (x - 5)(2x + 3)
2x^2 - 7x -15
=2x^2 - 10x +3x -15
=2x(x - 5) + 3(x - 5)
=(x - 5)(2x + 3)
(x-5)(2x+3)
Instead of coming for evry problem to Y!A, why not learn how to factorise a quadratic first?
2x² - 7x - 15
2x² - 10x + 3x - 15
2x ( x - 5 ) + 3 ( x - 5 )
( 2x + 3 ) ( x - 5 )
(2x+3)(x-5)