Re-write the quadratic function below in Standard Form y=5(x-5)(x+3) Answer: y=

Expand Quadratic Function: Expand the quadratic function using the distributive property (also known as the FOIL method for binomials). We need to multiply the two binomials (x-5) and (x+3) together. y = 5(x-5)(x+3) = 5[(x)(x) + (x)(3) - (5)(x) - (5)(3)]Simplify Expression: Continue simplifying the expression by combining like terms. y = 5[x^2 + 3x - 5x - 15] = 5[x^2 - 2x - 15]Distribute Coefficients: Distribute the 5 across each term inside the brackets. y = 5(x^2) - 5(2x) - 5(15) = 5x^2 - 10x - 75Write in Standard Form: Write the final expression in standard form, which is Ax^2 + Bx + C. The standard form of the quadratic function is: y = 5x^2 - 10x - 75

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Re-write the quadratic function below in Standard Form

y=5(x-5)(x+3)
Answer:
y=

Re-write the quadratic function below in Standard Form $\newline$ $y=5(x-5)(x+3)$ $\newline$ 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=5(x-5)(x+3)

\newline

Answer:

y=

Expand Quadratic Function: Expand the quadratic function using the distributive property (also known as the FOIL method for binomials). $\newline$ We need to multiply the two binomials $(x-5)$ and $(x+3)$ together. $\newline$ $y = 5(x-5)(x+3)$ $\newline$ $= 5[(x)(x) + (x)(3) - (5)(x) - (5)(3)]$
Simplify Expression: Continue simplifying the expression by combining like terms. $\newline$ $y = 5[x^2 + 3x - 5x - 15]$ $\newline$ $= 5[x^2 - 2x - 15]$
Distribute Coefficients: Distribute the $5$ across each term inside the brackets. $y = 5(x^2) - 5(2x) - 5(15)$ $= 5x^2 - 10x - 75$
Write in Standard Form: Write the final expression in standard form, which is $Ax^2 + Bx + C$ . The standard form of the quadratic function is: $y = 5x^2 - 10x - 75$