# Find a quadratic polynomial whose sum and product respectively of

Find a quadratic polynomial whose sum and product respectively of the zeroes are as given also find the zeoes of these polynomial by factorization
(-8/3),(4/3)
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Durst37
Step 1 Given
$\left(\frac{-8}{3}\right),\frac{4}{3}$
Step 2 Finding the zeroes
Sum of the zeroes $=\frac{-8}{3}$
Product of the zeroes $=\frac{4}{3}$

Then,
$p\left(x\right)={x}^{2}-\frac{8x}{3}+\frac{4}{3}$
$p\left(x\right)=3{x}^{2}-8x+4$
By factorization
$3{x}^{2}-8x+4=0$
$3{x}^{2}-\left(6x+2x\right)+4=0$
$3{x}^{2}-6x-2x+4=0$
3x(x-2)-2(x-2)=0
(x-2)(3x-2)=0
x-2=0; 3x-2=0
$x=2;x=\frac{2}{3}$
$\therefore x=2;\frac{2}{3}$
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Maricela Alarcon
Step 1
Let the roots of a quadratic equation
$a{x}^{2}+bx=c=0$ be
So, $\alpha +\beta =\frac{-b}{a}=\frac{-8}{3}$...(1)
$\alpha \beta =\frac{c}{a}=\frac{4}{3}$...(2)
$a{x}^{2}+bx+c=0$
$a\left({x}^{2}+\frac{b}{a}x+\frac{c}{a}\right)=0$
Step 2
${x}^{2}-\left(\frac{-b}{a}\right)x+\frac{c}{a}=0$
${x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta =0$
Substitute (1) and (2) in above
${x}^{2}-\left(\frac{-8}{3}\right)x+\frac{4}{3}=0$
$3{x}^{2}+8x+4=0$
$\therefore$ The quadratic equation is $3{x}^{2}+8x+4=0$
Step 3
$3{x}^{2}+8x+4=0$
$3{x}^{2}+2x+6x+4=0$
x(3x+2)+2(3x+2)=0
(3x+2)(x+2)=0
$x=\frac{-2}{3},x=-2$
$\therefore$ The roots are $\frac{-2}{3},-2$
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