Question

plzz.... help me.....​

Answer 1

$\huge\underline\mathfrak{Answer-}$

All the zeroes are : 3√2, -3√2, $\dfrac{-1}{4}$, 0

$\huge\underline\mathfrak{Explanation-}$

Given zeroes : 3√2 and -3√2

Sum of zeroes : 3√2 + ( - 3√2 ) = 0

Product of zeroes : 3√2 × ( - 3√2 ) = -9 × 2 = -18

For making Quadratic polynomial :

$\boxed{x^2-Sx+P}$

Here, Sx = sum of zeroes and p = product of zeroes

required polynomial : x² - 0x + ( -18 ) = x² - 18

So, we have to divide 4x⁴ + x³ - 72x -18 by x² - 18.

*Refer to the attachment for division*

For finding other zeroes :

4x² + x = 0

By taking x as common,

$\implies$ x ( 4x + 1 ) = 0

$\implies$ x = 0 and 4x + 1 = 0

$\implies$ x = 0 and x = $\dfrac{-1}{4}$

Therefore, all the zeroes of this polynomial are : 3√2, -3√2, 0 and $\bold{\dfrac{-1}{4}}$

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