Let A and B be the remainders when the polynomials y3+2y2−5ay−7 and y3+ay2−12y+6 are divided by y+1 and y−2 respectively. If 2A+B=6, find the value of a.
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Solution
y3+2y2−5ay−7 leaves a remainder A, when divided by y+1.
So, plugging in y=−1 in the above polynomial, we can find remainder A.
A=−1+2+5a−7=5a−6
y3+2y2−5ay−7 leaves a remainder B, when divided by y−2.
So, plugging in y=2 in the above polynomial, we can find remainder A.