Let R1 and R2 be the remainders
when polynomials x^3 + 2x^2 - 5ax
- 7 and x^ 3 + ax^2 - 12 x + 6 are
divided by ( x + 1 ) and ( x - 2 )
respectively. If 2R1 + R2 = 6, find
a. Please work the complete solution.
Got you wrong for this time, (x+1) and (x+2) aren't factors, they leave remainder R1 and R2 stated clearly at the top, by the way, Thanks for help, I've solved this question and the value of a is 2.
Its the bad way to do the Question. I think you should try another.🤔
To find the value of 'a', we need to use the Remainder Theorem and the given conditions.
The Remainder Theorem states that if a polynomial f(x) is divided by (x - c), then the remainder is equal to f(c).
Given that R1 is the remainder when the polynomial x^3 + 2x^2 - 5ax - 7 is divided by (x + 1), we can apply the Remainder Theorem. So, substituting x = -1 into the polynomial, we get:
R1 = (-1)^3 + 2(-1)^2 - 5a(-1) - 7
R1 = -1 + 2 + 5a - 7
R1 = 5a - 6
Similarly, R2 is the remainder when the polynomial x^3 + ax^2 - 12x + 6 is divided by (x - 2). So, substituting x = 2 into the polynomial, we get:
R2 = (2)^3 + a(2)^2 - 12(2) + 6
R2 = 8 + 4a - 24 + 6
R2 = 4a - 10
Now, we are given that 2R1 + R2 = 6. Substituting the expressions for R1 and R2 derived earlier, we can solve for 'a':
2(5a - 6) + (4a - 10) = 6
10a - 12 + 4a - 10 = 6
14a - 22 = 6
14a = 6 + 22
14a = 28
a = 28/14
a = 2
Therefore, the value of 'a' is 2.
Actually you solved it correctly but its 2R1, so you need to multiply your R1 by 2 and then equate it by 6. Little mistake but it won't fetch you marks in examinations.
if x+1 is a factor, then f(-1) = 0 from the first function:
f(-1) = -1 + 2 -5a(-1) - 7 = R1
R1 = 5a -6
if x-2 is a factor of the second function
f(2) = 8 + 4a - 24 + 6
= 4a - 10 = rR2
R1 + R2 = 6
5a-6 + 4a-10 = 6
9a = 22
a=22/9