Question

One zero of the polynomial $3x^3+16x^2+15x-18$ is $\frac{2}{3}$. Find the other zeros of the polynomial.

Answer 1

$p\left(x\right)=3x^3+16x^2+15x-18$

$x=\frac{2}{3}$

$x-\frac{2}{3}=0$

$\frac{3x-2}{3}=0$

$\Rightarrow 3x-2=0$

$x^2+6x+9$

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$3x-2)3x^3+16x^2+15x-18$

$3x^2-2x^2$

$(-)$ $(+)$

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$18x^2+15x-18$

$18x^2-12x$

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$27x-18$

$27x-18$

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$0$

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$x^2+6x+9$

$x^2+3x+3x+9$

$x(x+3)+3(x+3)$

$(x+3)(x+3)$

zeroes

$x+3=0$ $x+3=0$

$x=-3$ $x=-3$

Zeroes of $p\left(x\right)=-3,-3,\frac{2}{3}$.