If x3 + ax2 – bx + 10 is divisible by x2 – 3x + 2,
find the values of
1) a-b
2) 2a-b

A=2 and B=13
Explanation: The Factor Theorem states that if a is the root of any polynomial p(x) that is if p(a)=0, then (x−a) is the factor of the polynomial p(x).

Let p(x)=x
3
+ax
2
−bx+10 and g(x)=x
2
−3x+2
Factorise g(x)=x
2
−3x+2:
x
2
−3x+2=x
2
−2x−x+2=x(x−2)−1(x−2)=(x−2)(x−1)
Therefore, g(x)=(x−2)(x−1)
It is given that p(x) is divisible by g(x), therefore, by factor theorem p(2)=0 and p(1)=0. Let us first find p(2) and p(1) as follows:
p(1)=1
3
+(a×1
2
)−(b×1)+10=1+(a×1)−b+10=a−b+11
p(2)=2
3
+(a×2
2
)−(b×2)+10=8+(a×4)−2b+10=4a−2b+18
Now equate p(2)=0 and p(1)=0 as shown below:
a−b+11=0
⇒a−b=−11(1)
4a−2b+18=0
⇒2(2a−b+9)=0
⇒2a−b+9=0
⇒2a−b=−9(2)
Now subtract equation 1 from equation 2:

(2a−a)+(−b+b)=(−9+11)
⇒a=2
Substitute a=2 in equation 1:
2−b=−11
⇒−b=−11−2
⇒−b=−13
⇒b=13
Hence, a=2 and b=13.

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3