Find the equation of the parabola that has zeros of x = –2 and x = 3 and a y–intercept of (0,–30).

Mathematics

1 answer:

Anit [1.1K]3 years ago

4 0

Answer:

Removing answer.

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Step-by-step explanation:

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Multiply. (2x2 + 4x - 3)(x2 - 2x + 5)

BigorU [14]

For a moment, let $y=x^2-2x+5$ . Then by the distributive property,

$(2x^2+4x-3)p=2x^2p+4xp-3p$

Again by the distributive property,

$2x^2p=2x^2(x^2-2x+5)=2x^4-4x^3+10x^2$

$4xp=4x(x^2-2x+5)=4x^3-8x^2+20x$

$-3p=-3(x^2-2x+5)=-3x^2+6x-15$

Taking everything together, we get

$(2x^2+4x-3)(x^2-2x+5)=(2x^4-4x^3+10x^2)+(4x^3-8x^2+20x)+(-3x^2+6x-15)$
$(2x^2+4x-3)(x^2-2x+5)=2x^4-x^2+26x-15$