38) The polynomials p(x) = x^3 + 2x^2 – 5ax - 8 and q(x) = x^3 + ax^2 –12x -11, when divided by (x – 2) and (x – 3) respectively leave the remainders r1 and r2. If r2 – r1 = 10, find the value of a.
Answers
Step-by-step explanation:
Answer
The Remainder Theorem states that when we divide a polynomial p(x) by any factor (x−a); which is not necessarily a factor of the polynomial; we will obtain a new smaller polynomial and a remainder, and this remainder is the value of p(x) at x=a, that is p(a).
Let p(x)=2x3−5x2+x+a and q(x)=ax3+2x2−3 and the factor given is g(x)=x−2, therefore, by remainder theorem, the remainders are p(2) and q(2) respectively and thus,
p(2)=(2×23)−(5×22)+2+a=(2×8)−(5×4)+2+a=16−20+2+a=a−2q(2)=(a×23)+(2×22)−3=(a×8)+(2×4)−3=8a+8−3=8a+5
Now since it is given that both the polynomials p(x)=2x3−5x2+x+a and q(x)=ax3+2x2−3 leave the remainders R1 and R2Answer
The Remainder Theorem states that when we divide a polynomial p(x) by any factor (x−a); which is not necessarily a factor of the polynomial; we will obtain a new smaller polynomial and a remainder, and this remainder is the value of p(x) at x=a, that is p(a).
Let p(x)=2x3−5x2+x+a and q(x)=ax3+2x2−3 and the factor given is g(x)=x−2, therefore, by remainder theorem, the remainders are p(2) and q(2) respectively and thus,
p(2)=(2×23)−(5×22)+2+a=(2×8)−(5×4)+2+a=16−20+2+a=a−2q(2)=(a×23)+(2×22)−3=(a×8)+(2×4)−3=8a+8−3=8a+5
Now since it is given that both the polynomials p(x)=2x3−5x2+x+a and q(x)=ax3+2x2−3 leave the remainders R1 and R2