Question

Can u help me ASAP. i need to know how to do it step by step​

Answer 1

Answer:

19

Step-by-step explanation:

$f(x) =4x^3-8x^2+ax+b$ has a factor $2x-1$ and

when divided by $x+2$, remainder is 20.

To find:Remainder when divided by (x-1)  ?

Solution:

$2x-1$ is a factor

$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ when we put this value of x to the function, it will become 0.

i.e.

$\Rightarrow f(\dfrac{1}{2}) =0 =4\times (\dfrac{1}2)^3-8(\dfrac{1}2)^2+\dfrac{a}{2}+b=0\\\Rightarrow \dfrac{1}{2}-2+\dfrac{a}{2}+b=0\\\Rightarrow 1-4+a+2b=0\\\Rightarrow a +2b=3 ......(1)$

Remainder is 20 when f(x) is divided by $x+2$

i.e.

$f(-2) =20$

$\Rightarrow f(-2) =20 =4\times (-2)^3-8(-2)^2-2a+b=20\\\Rightarrow -32-32-2a+b=20\\\Rightarrow -2a+b=84 ...... (2)$

Solving (1) and (2), Multiply equation (1) by 2 and adding to (2):

$5b=6+84\\\Rightarrow b = \dfrac{90}{5} = \bold{18}$

By equation (1):

$a+2(18) = 3\\\Rightarrow a = -33$

So, the equation becomes:

$f(x) =4x^3-8x^2-33x+18$

$\Rightarrow f(1) = 4(1) -8 (1) -33(1) +18 = \bold{19}$

So, when divided by (x-1), remainder will be 19.