Question

If the remainder on division of x³+2x²+kx+7 by x-3 is 25 find the value of k and the quotient and find the zeroes of the quotient

Answer 1

Answer:

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.

Answer 2

Answer: k=-9, zeroes are -2,-3,3

Step-by-step explanation:

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,

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.