Question

the polynomial kx³+3x²-8 and 3x³-5x+8 are divided by X+2.if the remainder in each case is the same ,find the value of k​

Answer 1

Answer:

$\frac{5}{4}$

Step-by-step explanation:

using remainder theorem,

$k(-2)^{3}+3(-2)^{2}-8=3(-2)^{3}-5(-2)+8\\\\12-8-8-10+24=8k\\k=\frac{10}{8}=\frac{5}{4}$

Answer 2

$\huge \rm {Answer:-}$

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$\large \to \tt {\underline{Given:-}}$

★The two polynomials,

$\implies {kx^{3}+3x^{2}-8}$

★And,

$\implies {3x^{3}-5x+8}$

★Leave the same remainder when divided by x+2

$\large \implies {x+2=0}$

$\large \implies \pink {x=-2}$

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$\large \to \tt {\underline{To\: find:-}}$

★ The value of "K"

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★Substituting,x=-2 in the above given polynomials-

$\large \to \tt {\underline{Polynomial\: 1:-}}$

$\implies {P(x)=kx^{3}+3x^{2}-8}$

$\implies {P(-2)=k(-2)^{3}+3(-2)^{2}-8}$

$\implies {P(-2)=k(-8)+3(4)-8}$

$\implies {P(-2)=-8k+12-8}$

$\to \purple {\fbox {P(-2)=-8k+4}}$

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$\large \to \tt {\underline{Polynomial\: 2:-}}$

$\implies {Q(x)=3x^{3}-5x+8}$

$\implies {Q(-2)=3(-2)^{3}-5(-2)+8}$

$\implies {Q(-2)=3(-8)+10+8}$

$\implies {Q(-2)=-24+10+8}$

$\to \red {\fbox {Q(-2)=-6}}$

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★As the two polynomials leave the same remainder when divided,they can be equated .

$\implies {P(-2)=Q(-2)}$

$\implies {-8k+4=-6}$

$\implies {-8k=-6-4}$

$\implies {-8k=-10}$

$\large \implies {k=\frac{\cancel{-}10}{\cancel{-}8}}$

$\large \to \orange {k=\frac{5}{4}}$

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$\large \to \tt {\underline{Remainder\: of\: the\: polynomials:-}}$

$\small\to \tt {Polynomial\: 2=-6}$

$\large \to \tt {\underline{Polynomial\: 1:-}}$

$\implies {x=-2\: and\: k=\frac{5}{4}}$

$\implies {kx^{3}+3x^{2}-8}$

$\large \implies {\frac{5}{4}(-2)^{3}+3(-2)^{2}-8}$

$\implies {\frac{5}{4}-8+3(4)-8}$

$\implies {5\times-2+12-8}$

$\implies {-10+12-8}$

$\large \to \blue {\fbox {-6}}$

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$\large \to \tt {\underline{Thereupon, verified✓}}$

$\implies \green {★The \: remainder=-6}$