Question

BRAINLIEST yr answer MARKEd​

Answer 1

Here,

$\longrightarrow p(x)=2x^4+7x^3-19x^2-14x+30$

If $\dfrac{3}{2}$ and $-5$ are zeroes of $p(x),$ then $(2x-3)$ and $(x+5)$ are factors of $p(x).$

Product of these factors is,

$\longrightarrow(2x-3)(x+5)=2x^2+7x-15$

We've to divide $p(x)$ by this product.

$\longrightarrow p(x)=2x^4+7x^3-19x^2-14x+30$

$\longrightarrow p(x)=2x^4+7x^3+(-15-4)x^2-14x+30$

$\longrightarrow p(x)=2x^4+7x^3-15x^2-4x^2-14x+30$

$\longrightarrow p(x)=x^2(2x^2+7x-15)-2(2x^2+7x-15)$

$\longrightarrow p(x)=(2x^2+7x-15)(x^2-2)$

This means $p(x)$ divided by $(2x^2+7x-15)$ gives quotient $(x^2-2).$

This quotient is product of other factors of $p(x).$

$\longrightarrow x^2-2=\left(x+\sqrt2\right)\left(x-\sqrt2\right)$

So other factors of $p(x)$ are $\left(x+\sqrt2\right)$ and $\left(x-\sqrt2\right).$

Hence other zeroes are $\bf{-\sqrt2}$ and $\bf{\sqrt2}\,.$