Find the Quadratic Equation Given the Roots -3 and 15

$- 3$ and $15$

$x = - 3$ and $x = 15$ are the two real distinct solutions for the quadratic equation, which means that $x + 3$ and $x - 15$ are the factors of the quadratic equation.

$(x + 3) (x - 15) = 0$

Expand $(x + 3) (x - 15)$ using the FOIL Method.

Apply the distributive property.

$x (x - 15) + 3 (x - 15) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 15 + 3 (x - 15) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 15 + 3 x + 3 \cdot - 15 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 15 + 3 x + 3 \cdot - 15 = 0$

Move $- 15$ to the left of $x$ .

$x^{2} - 15 \cdot x + 3 x + 3 \cdot - 15 = 0$

Multiply $3$ by $- 15$ .

$x^{2} - 15 x + 3 x - 45 = 0$

Add $- 15 x$ and $3 x$ .

$x^{2} - 12 x - 45 = 0$

The standard quadratic equation using the given set of solutions ${- 3, 15}$ is $y = x^{2} - 12 x - 45$ .

$y = x^{2} - 12 x - 45$

Find the Quadratic Equation Given the Roots -3 and 15

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