Re-write the quadratic function below in Standard Form y=4(x+2)(x+7) Answer: y=

Expand Quadratic Function: Expand the quadratic function using the distributive property (also known as FOIL - First, Outer, Inner, Last). We need to multiply the terms (x+2) and (x+7) together and then multiply the result by 4. y = 4(x+2)(x+7) First, we'll multiply (x+2) by (x+7). (x+2)(x+7) = x(x) + x(7) + 2(x) + 2(7) = x^2 + 7x + 2x + 14 = x^2 + 9x + 14Multiply Expanded Binomial: Multiply the expanded binomial by 4. Now we multiply each term in the binomial x^2 + 9x + 14 by 4. y = 4(x^2 + 9x + 14) = 4x^2 + 36x + 56

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Re-write the quadratic function below in Standard Form

y=4(x+2)(x+7)
Answer:
y=

Re-write the quadratic function below in Standard Form $\newline$ $y=4(x+2)(x+7)$ $\newline$ Answer: $y=$

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Algebra 1

Multiply two binomials

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Q. Re-write the quadratic function below in Standard Form

\newline

y=4(x+2)(x+7)

\newline

Answer:

y=

Expand Quadratic Function: Expand the quadratic function using the distributive property (also known as FOIL - First, Outer, Inner, Last). $\newline$ We need to multiply the terms $(x+2)$ and $(x+7)$ together and then multiply the result by $4$ . $\newline$ $y = 4(x+2)(x+7)$ $\newline$ First, we'll multiply $(x+2)$ by $(x+7)$ . $\newline$ $(x+2)(x+7) = x(x) + x(7) + 2(x) + 2(7)$ $\newline$ $= x^2 + 7x + 2x + 14$ $\newline$ $= x^2 + 9x + 14$
Multiply Expanded Binomial: Multiply the expanded binomial by $4$ . $\newline$ Now we multiply each term in the binomial $x^2 + 9x + 14$ by $4$ . $\newline$ $y = 4(x^2 + 9x + 14)$ $\newline$ $= 4x^2 + 36x + 56$