if the polynomials 2x³+ax²+3x-5and x³+x²-4x+a leave same remainder when divided by x-2,find the value of a
Answers
The Remainder Theorem states that when you divide a polynomial p(x) by any factor (x−a); which is not necessarily a factor of the polynomial; you 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)=2x
3
+ax
2
+3x−5 and q(x)=x
3
+x
2
−4x−a and the factor given is g(x)=x−1, therefore, by remainder theorem, the remainders are p(1) and q(1) respectively and thus,
p(1)=(2×1
3
)+(a×1
2
)+(3×1)−5=(2×1)+(a×1)+3−5=2+a−2=a
q(1)=1
3
+1
2
−(4×1)−a=1+1−4−a=−2−a
Now since it is given that both the polynomials p(x)=2x
3
+ax
2
+3x−5 and q(x)=x
3
+x
2
−4x−a leave the same remainder when divided by (x−1), therefore p(1)=q(1) that is:
a=−a−2
⇒a+a=−2
⇒2a=−2
⇒a=−
2
2
⇒a=−1
Hence, a=−1.
x-2=0
x=2
ATQ
2x³+ax²+3x-5=x³+x²-4x+a
2(2)³+a(2)²+3(2)-5=(2)³+(2)²-4(2)+a
16+4a+6-5=8+4-8+a
17+4a=4+a
4a-a= -17+4
3a= -13
a= -13/3