Question

If A and B be the remainders when the polynomials x^3+2x^2-5ax-7 and x^3+ax^2-12x+6 are divided by (x+1) and (x-2).Respectively and 2A+B=6 . Find the value of A.​

Answer 1

Answer:

Step1:

p(x)=x3+2x2−5ax−7Since p(x) is divisible by (x+1) the remainder is p(−1)p(−1)=−1+2+5a−7=5a−6=A

Step2:

t(x)=x3+ax2−12x+6Since t(x) is divisible by (x−2) the remainder is t(2)t(2)=(2)^3+4a−24+6=4a−10=B

Step3:

Since 2A+B=6we know that A=(5a−6),B=4a−10ATherefore 2A+B=6⇒2(5a−6)+4a−10=6Solving for ′a10a−12+4a−10=610a−12+4a−10=614a−22=614a=28a = 28 ÷ 14 We get a=2

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Answer 2

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Let p(x)=x³+2x²−5ax−7

and q(x)=x³+ax²−12x+6 be the given polynomials,

Now, R¹= Remainder when p(x) is divided by x+1.

⇒R¹=p(−1)

⇒R¹=(−1)³+2(−1)²−5a(−1)−7[∵p(x)=x²+2x²−5ax−7]

⇒R¹=−1+2+5a−7

⇒R¹=5a−6

And R² = Remainder when q(x) is divided by x-2

⇒R¹ =q(2)

⇒R²=(2) 3+a×2²−12×2+6[∵q(x)=x²+ax²−12x−6]

⇒R²=8+4a−24+6

⇒R²=4a−10

Substituting the values of R¹ and R² in 2R¹+

R² =6, we get

⇒2(5a−6)+(4a−10)=6

⇒10a−12+4a−10=6

⇒14a−22=6

⇒14a−28=0

⇒a=2

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