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I'm assuming you want to solve the equation
2x2 - 12x - 80 = 0
Divide both sides by 2:
x2 - 6x - 40 =0
What numbers multiply to get -40 and add to give you -6?
Factors of 40:
(2,20)
(4,10)
(5,8)
Thus, the factors of -40 are the same, but with either number in the pair being negative. Which two have a difference of 6? (4,10). Since we want -6, we should choose (4,-10)
Thus we can factor the equation to get:
(x + 4)(x - 10) = 0
The two solutions are then x= -4, x=10
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Can 2x2 -12x- 80 be factored?
Yes. You must use the "slip and slide" factoring method because there is a 2 coefficient infront of x2. 2x2-12x-80 = (x-10)(2x+8)
what is the factor the polynomial 2x2-12x-80?
2(x + 4)(x - 10)
2x3 plus 2x2 - 12x?
Do you want its factorisation? 2x3 + 2x2 - 12x = 2x(x2 + x - 6) = 2x(x + 3)(x - 2).
What is the factorization of the polynomial below -2x3-2x2 plus 12x?
(-2x3 - 2x2 + 12x) = -2x (x2 + x - 6) = -2x (x + 3) (x - 2)
Can 2x2-12x-80 be factored?
Yes because dividing all terms by 2 it becomes x2-6x-40 = (x-10)(x+4) when factored
