Question #9c3e2

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Question #9c3e2

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Answer 1 · 2017-02-14T11:42:17

${- 3, - \frac{1}{2}}$ ${-3,-1/2}$

Explanation:

If $\frac{1}{2}$ $1/2$ and $3$ $3$ are zeroes then $y = 4 x^{4} - 37 x^{2} + 9 = p_{2} (x) (x - \frac{1}{2}) (x - 3)$ $y=4x^4-37x^2+9=p_2(x)(x-1/2)(x-3)$ where $p_{2} (x) = a x^{2} + b x + c$ $p_2(x)=a x^2+bx+c$ is a generic order $2$ $2$ polynomial.

Substituting and equating coefficients we have

$4 x^{4} - 37 x^{2} + 9 = (a x^{2} + b x + c) (x - \frac{1}{2}) (x - 3)$ $4x^4-37x^2+9=(ax^2+bx+c)(x-1/2)(x-3)$

so

${((3 c)/2=9),(-(3 b)/2 + (7 c)/2=0),( - (3 a)/2 + (7 b)/2 - c=37),((7 a)/2 - b=0),(a=4):}$

now solving for $a,b,c$ we obtain

$a = 4, b = 14, c = 6$

so $p_2(x)=4x^2+14x+6$ with roots

$x={-3,-1/2}$ so finally

$y=4x^4-37x^2+9=4(x^2-3^2)(x^2-(1/2)^2)$

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Question #9c3e2

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