Question

The polynomials kx3+3x2-8 and 3x​3-5x+k divided by x+2, if the remainder in each case is the same find the value of k.

Answer 1

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here is your answer,

⇒x+2=0

⇒x=0-2

⇒x=-2

kx³+3x²-8

⇒ k * (-2)³ + 3* (-2)² -8=0

⇒ k * -8 + 3*4 + -8=0

⇒ -8k + 12-8=0

⇒ -8k + 4 =0

⇒-8k= 0-4

⇒ -8k= -4

⇒k= -4/-8

k= 1/2

3x³-5x+k

⇒ 3*(-2)³ - 5 *-2 +k=0

⇒ 3*-8 + 10 +k=0

⇒ -24+ 10 +k =0

⇒ -14 +k=0

⇒k=14

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hope it helps

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