Question

if x is equal to x cube minus 2 X square + kx + 5 is divided by x minus 2 the remainder is 11 find the ke hence find all the zeros of x cube + 2 X square + 3 X + 1

Answer 1

plz check this attachment
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Answer 2

$Given \:p(x) = x^{3} -2x^{2}+kx+5$

$If \: p(x) \: divided \:by \:(x-2) \: then \:the \\remainder \:is \: p(2)$

$But , \: p(2) = 11 \: (given)$

$\implies 2^{3} -2\times 2^{2} + k \times 2 + 5 = 11$

$\implies 8 - 8 + 2k = 11 - 5$

$\implies 2k = 6$

$\implies k = \frac{6}{2}$

$\implies k = 3\:--(1)$

$ii ) Given \: p(x) = x^{3} +kx^{2}+3x+1$

$p(x) = x^{3} +3x^{2}+3x+1\: (From \:(1) ]$

/* put x = -1 , we get */

$p(-1) = (-1)^{3} + 3(-1)^{2} + 3(-1) + 1 \\= -1 + 3 - 3 + 1 \\= 0$

$\therefore (x+1) \:is \:a \:factor \:of \:p(x)$

x+1) x³+3x²+3x+1 ( x²+2x +1

*****x³ + x²

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*********2x² + 3x

********* 2x² + 2x

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************** x + 1

************** x + 1

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Remainder (0)

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$Now, p(x) = (x+1)(x^{2} + 2x +1 )\\= (x+1)(x+1)^{2}\\= (x+1)(x+1)(x+1)$

Therefore.,

$\red { Factors \: of \: p(x) } \green{= (x+1)(x+1)(x+1)}$

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