What is the standard form of $y = (x + 4) (2 x - 3) - 3 x^{2}$ $y=(x+4)(2x-3) -3x^2$ ?

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Answer 1 · 2015-12-05T23:55:40

The standard form is $y = - x^{2} + 5 x - 12$ $y=-x^2+5x-12$ .

$y = (x + 4) (2 x - 3) - 3 x^{2}$ $y=(x+4)(2x-3)-3x^2$

The standard form for a quadratic equation is $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a, b, and c$ $a, b, and c$ are coefficients.

First Foil the two binomials.

![http://hubpages.com/education/Using-the-FOIL-Method-to-Expand-Products]( useruploads.socratic.org )

$(x + 4) (2 x - 3)$ $(x+4)(2x-3)$

$a = x, b = 4, c = 2 x, d = - 3$ $a=x, b=4, c=2x, d=-3$

$(x + 4) (2 x - 3) = a c + a d + b c + b d$ $(x+4)(2x-3)=ac+ad+bc+bd$

$(x + 4) (2 x - 3) = (x \cdot 2 x) + (x \cdot - 3) + (4 \cdot 2 x) + (4 \cdot - 3)$ $(x+4)(2x-3)=(x*2x)+(x*-3)+(4*2x)+(4*-3)$

$(x + 4) (2 x - 3) = (2 x^{2}) + (- 3 x) + (8 x) + (- 12)$ $(x+4)(2x-3)=(2x^2)+(-3x)+(8x)+(-12)$

Combine like terms.

$(x + 4) (2 x - 3) = 2 x^{2} + 5 x - 12$ $(x+4)(2x-3)=2x^2+5x-12$

Return to original equation, keeping the foiled results.

$y = 2 x^{2} + 5 x - 12 - 3 x^{2}$ $y=2x^2+5x-12-3x^2$

Combine like terms.

$y = 2 x^{2} - 3 x^{2} + 5 x - 12$ $y=2x^2-3x^2+5x-12$

$y = - x^{2} + 5 x - 12$ $y=-x^2+5x-12$

What is the standard form of y=(x+4)(2x-3) -3x^2y=(x+4)(2x−3)−3x2y=(x+4)(2x-3) -3x^2?