Question

Any Real Genius Here??
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1⃣If the remainder division of
${x}^{3} + 2 {x}^{2} + kx + 3$
by
$x - 3$
is 21,Find the quotient and value of k
Hence find the zeros of cubic polynomial
${x}^{3} + 2 {x}^{2} + kx - 18$
Best Answer Will be mark as brainliest❤️❤️❤️

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Answer 1

★Heya★

Divedent = Divisor × Q + Remainder.

Here

Divedent = x³ + 2x² + kx + 3

Divisor = ( x - 3 )

Remainder = 21

Let the Quotient be = Q

( x - 3 ) is Divisor so, x = 3 is the root of given Equation.

=>

( x³ + 2x² + kx + 3 ) = ( x - 3 ) × Q + 21

put x = 3

27 + 18 + 3k + 3 = 21

=>

k = -9

Now the Equation is

x³ + 2x² + 3x + 3 = ( x - 3 ) ( x² + 5x + 6 )

=>

( x - 3 ) × ( x² + 5x + 6 ) = 0

=>

( x - 3 ) = 0

OR

( x ² + 5x + 6 ) = 0

=>

x = 3

OR

x = {-5 ±√(25 - 24 )}/2

=>

x = (-5 ± 1)/2

=>

x = -3 OR x = -2

So, the roots of Equation are

3 , -3 and -2