Answer:
19
Step-by-step explanation:
[tex]f(x) =4x^3-8x^2+ax+b[/tex] has a factor [tex]2x-1[/tex] and
when divided by [tex]x+2[/tex], remainder is 20.
To find:Remainder when divided by (x-1) ?
Solution:
[tex]2x-1[/tex] is a factor
[tex]2x - 1 = 0 \Rightarrow x = \frac{1}{2}[/tex] when we put this value of x to the function, it will become 0.
i.e.
[tex]\Rightarrow f(\dfrac{1}{2}) =0 =4\times (\dfrac{1}2)^3-8(\dfrac{1}2)^2+\dfrac{a}{2}+b=0\\\Rightarrow \dfrac{1}{2}-2+\dfrac{a}{2}+b=0\\\Rightarrow 1-4+a+2b=0\\\Rightarrow a +2b=3 ......(1)[/tex]
Remainder is 20 when f(x) is divided by [tex]x+2[/tex]
i.e.
[tex]f(-2) =20[/tex]
[tex]\Rightarrow f(-2) =20 =4\times (-2)^3-8(-2)^2-2a+b=20\\\Rightarrow -32-32-2a+b=20\\\Rightarrow -2a+b=84 ...... (2)[/tex]
Solving (1) and (2), Multiply equation (1) by 2 and adding to (2):
[tex]5b=6+84\\\Rightarrow b = \dfrac{90}{5} = \bold{18}[/tex]
By equation (1):
[tex]a+2(18) = 3\\\Rightarrow a = -33[/tex]
So, the equation becomes:
[tex]f(x) =4x^3-8x^2-33x+18[/tex]
[tex]\Rightarrow f(1) = 4(1) -8 (1) -33(1) +18 = \bold{19}[/tex]
So, when divided by (x-1), remainder will be 19.
