Question

If the remainder on division of x 3 + 2x 2 + kx +3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x 3 + 2x 2 + kx – 18

Answer 1

We know that,
Dividend=Divisor×Quotient+Remainder
Given that,
Dividend=x³+2x²+kx+3
Divisor=x-3
Remainder=21
Now we have,
x³+2x²+kx+3=(x-3)quotient+21...............1

(x³+2x²+kx-18)/(x-3)=quotient
Now the remainder will be equal to 0.

From above picture we get,
Remainder=0
3(k+15)-18=0

k+15=6

k=-9
Now eq1 we have,
x³+2x²-9x+3=(x-3)(x²+5x+6)+21

x³+2x²-9x-18=(x-3)(x²+2x+3x+6)

x³+2x²-9x-18=(x-3){x(x+2)+3(x+2)}

x³+2x²-9x-18=(x-3)(x+3)(x+2)

Therefore zeroes are 3 , -3 and -2


Hence value of k=-9
and zeroes of given cubic polynomial are 3 , -3 and -2.

Hope it helps!!

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