Question

When 2 x³+ 5 x²+ kx - 4 is divided by (x–2) the remainder is k .Find the value of the constant k


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Answer 1

Let x = 2

and p(x) = 2 x³+ 5 x²+ kx - 4

Putting, x = 2 in p(x) = 2 x³+ 5 x²+ kx - 4

$\implies$p(2) = 2(2)³+5(2)²+2k-4

$\implies$p(2) = 16+20+2k-4

$\implies$p(2)=32+2k

We know that p(x)=0

$\implies$2k+32=0

$\implies$2k=-32

$\implies\sf{k=\frac{-32}{2}}$

$\implies{\boxed{\sf{\purple{k=-16}}}}$

Hence, -16 is the value of the constant k.

Answer 2

Left f (x) = x³ +2x² - kx +4

x =2 = 0 ◗ x = 2

on dividing f (x) by x-2,, it leaves a reminder k

f (2) =k

$\tt(2)³+2(2)²-k(2)+4=k$

$\tt8+8-2k+4=k$

$\tt \: 20=3k$

$\tt \: k = \frac{20}{3} = 6 \frac{2}{3}$

I hope it will help you