Question

Match each quadratic function given in factored form
with its equivalent standard form listed on the left.
f(x) = (x + 2)(x – 6)
f(x) = (x – 4)(x+3)
f(x) = (x – 12)(x+1)
f(x) = (x – 3)(x +4)

Answer 1

Answer:

(x+2)(x-6)=x^2 - 4x -12

(x-4)(x+3)=x^2 - x - 12

(x-12)(x+1)=x^2 - 11x - 12

(x-3)(x+4)=x^2 + x - 12

Step-by-step explanation:

You do the calculations:

(x + a) (x + b) = x^2 +ax + bx + ab = x^2 + (a+b) + ab, where a and b are real numbers. (At least in this case)

Answer 2

The factored form of the expression as shown are $x^2 - x - 12, \ x^2 - 11x - 12, \ x^2 - 11x - 12 \ and \ x^2 + x - 12$

Expanding factored form of quadratic equation

Given the following functions;

$f(x) = (x + 2)(x -6) = x^2 - 6x + 2x - 12\\f(x)= x^2 - 4x - 12$

For the function'

$f(x) = (x - 4)(x+3) = x^2 + 3x - 4x - 12\\f(x) = x^2 - x - 12$

For the function f(x) = (x – 12)(x+1)

$f(x) = x^2 + x - 12x - 12\\f(x) = x^2 - 11x - 12$

For the function f(x) = (x – 3)(x +4)

$f(x) = (x -3)(x +4) =x^2 + 4x - 3x - 12\\f(x) = x^2 + x - 12$

Learn more on factoring here; https://brainly.com/question/25829061

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