Question

let a and b are remainders when the polynomials x3 +2x2-5ax-7 and x3+ax2-12x+6 are divided by x+1 and x-2 respectively . if 2a+b=6,find the value of A.

please solve this

Answer 1

Step 1:

p(x)=x3+2x2−5ax−7

Since p(x) is divisible by (x+1) the remainder is p(−1)

p(−1)=−1+2+5a−7

=5a−6

=A

Step 2:

t(x)=x3+ax2−12x+6

Since t(x) is divisible by (x−2) the remainder is t(2)

t(2)=(2)^3+4a−24+6

=4a−10

=B

Step 3:

Since 2A+B=6

we know that A=(5a−6),B=4a−10A

Therefore 2A+B=6⇒2(5a−6)+4a−10=6

Solving for ′a

10a−12+4a−10=610a−12+4a−10=6

14a−22=6

14a=28
a = 28 ÷ 14

We get a=2

Answer 2

Answer:

Step 1:

p(x)=x3+2x2−5ax−7

Since p(x) is divisible by (x+1) the remainder is p(−1)

p(−1)=−1+2+5a−7

=5a−6

=A

Step 2:

t(x)=x3+ax2−12x+6

Since t(x) is divisible by (x−2) the remainder is t(2)

t(2)=(2)^3+4a−24+6

=4a−10

=B

Step 3:

Since 2A+B=6

we know that A=(5a−6),B=4a−10A

Therefore 2A+B=6⇒2(5a−6)+4a−10=6

Solving for ′a

10a−12+4a−10=610a−12+4a−10=6

14a−22=6

14a=28

a = 28 ÷ 14

We get a=2

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