What multiplies to 4 and adds to -2?
I'm trying to find the zeroes of the function f(x)=x^2-2x-8. I did the AC/B tree (1*-8/-2) and from there I can't figure it out.
f(x) = x^2 - 2x - 8 = (x-1)^2 - 9
so, f(x) = 0 at x = 1±√9 = 4,-2
f(x) = (x-4)(x+2)
the real question was
What multiplies to -8 and adds to -2?
That is, which two factors of 8 differ by 2?
Y = x^2 - 2x - 8.
-8 = -4*2. Sum = -4 + 2 = -2 = B.
Y = (x-4)(x+2) = 0.
x-4 = 0, X = 4.
x + 2 = 0, X = - 2.
Solution set: (4,0), (-2, 0).
To find the zeroes of the function f(x) = x^2 - 2x - 8, you can use the factoring method or the quadratic formula. Let's first check if we can factor the equation.
You mentioned that you did the AC/B tree, which is a helpful tool for factoring quadratic equations. The AC/B tree helps you to find two numbers that multiply to give you the product of the coefficient of x^2 (A) and the constant term (C), and add up to the coefficient of x (B).
In this case, A = 1, B = -2, and C = -8. Your goal is to find two numbers that multiply to -8 and add up to -2.
To use the AC/B tree, factorize the constant term (-8) into pairs of numbers and check if any of them add up to -2:
-8 = 1 * (-8) = (-1) * 8 = 2 * (-4) = (-2) * 4.
Out of these pairs, (-2) and 4 add up to -2. Therefore, you can split the middle term (-2x) into -2x = 4x - 6x.
Now, rewrite the equation by splitting the middle term:
f(x) = x^2 + 4x - 6x - 8.
After grouping the terms, factorize by grouping:
f(x) = (x^2 + 4x) + (-6x - 8)
= x(x + 4) - 2(x + 4)
= (x + 4)(x - 2).
Now, you have factored the equation: f(x) = (x + 4)(x - 2).
To find the zeroes, set each factor equal to zero and solve for x:
(x + 4) = 0 or (x - 2) = 0.
For (x + 4) = 0: subtract 4 from both sides, and you get:
x = -4.
For (x - 2) = 0: add 2 to both sides, and you get:
x = 2.
Therefore, the zeroes of the function f(x) = x^2 - 2x - 8 are x = -4 and x = 2.