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Let R1 and R2 are 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. If R1+R2 = 6, find the value of a.

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Answer:

2.44

Explanation:

Given: x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6

Also, R1 + R2 = 6

in order to find the value of a:

Let p(x) = x³ + 2x² - 5ax - 7 and q(x) = x³ + ax² - 12x + 6

Using remainder theorem i.e if a polynomial p(x) is divisible by polynomial of form x - a then remainder is given by p(a).

Then,

R1 = p( -1 ) = (-1)³ + 2(-1)² - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6

R2 = q( 2 ) = 2³ + a(2)² - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10

Now,

R1 + R2 = 6

5a - 6 + 4a - 10 = 6

9a = 22

a=2.44

Therefore, Value of a is 2.44

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