What number should be subtracted from $4x^3+5x+3$ so that the resulting polynomial leaves remainder $-80$ when divided by $2x+5$?.
Let the required number to be subtracted be $K$.
let: $$P(x)=4x^3+5x+3-k$$
$$g(x)=2x+5=2(x+5/2)$$
Comparing $g(x)$ with $x-a $ , $a =\frac {-5}{2}$
Solving then gives $K=8$
but the answer in my book is $-152$. how?
Let us define $f(x) = 4x^3 +5x +3 -k$, where $k$ is the number that needs to be subtracted
By Remainder Theorem, $f(a)$ provides the remainder when $f(x)$ is divided by $(x-a)$
$$f(-\frac{5}{2}) = -72-k = -80$$ $$\therefore k = -72+80=8$$
I agree with you that the answer should be 8.
Let $f (x)=2x+5$ and $p (x)=4x^3+5x-3$. We have that: $$p (-\frac {5}{2}) -k = -80$$ $$\Rightarrow 4 (-\frac {5}{2})^3 +5 (-\frac {5}{2}) +3 - k = -80$$ $$-75+3-k = -80$$ $$\boxed {k = 8} $$ Hope it helps.
$$4x^3+5x+3=2x^2\underbrace{(2x+5)}-5x\underbrace{(2x+5)}+15\underbrace{(2x+5)}+3-75$$
(2x+5) is a factor.
So we have 2x+5=0
x=$\frac{-5}{2}$
When we put this into polynomial we get k = 8.
So your answer is correct.
