Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$

Question

Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?

From the problem I know that $f(4)=51$.

Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.

Then $$a+b+x(b-a)=0$$

I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.

Then $$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$

I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$

Answer 1

$f(x)=ax^3-ax^2+bx+4$

Since $f(4) = 51$, $51 =a(64-16)+4b+4 =48a+4b+4 $ so $12a+b =47/4 $.

Since $f(i) = 0$, $0 =a(-i+1)+ib+4 =i(b-a)+a+4 $ so $ a+4 = 0, a=-4, b-a = 0, b=a=-4 $.

Therefore $f(x) = -4x^3+4x^2-4x+4 $.

But this does not satisfy $f(4) = 51$.

Therefore the problem is wrong.

Answer 2

One more way to see the conditions imposed on $f(x)$ are inconsistent/impossible:

Since $x^2 + 1$ divides $f(x)$ with remainder $0$, $f(x)$ factors as

$ax^3 - ax^2 + bx + 4 = (x^2 + 1)(cx + d) = cx^3 + dx^2 + cx + d; \tag 1$

comparing coefficients:

$a = c = -d, \tag 2$

$b = c, \tag 3$

$ d = 4; \tag 4$

thus,

$a = b = c = -4; \tag 5$

thus,

$cx + d = -4x + 4, \tag 6$

and

$f(x) = -4x^3 + 4x^2 - 4x + 4; \tag 7$

then clearly $f(4)$ is even, so

$f(4) \ne 51. \tag 8$

If we choose to ignore the condition

$f(4) = 51, \tag 9$

we may still salvage the inference

$a + b = -8. \tag{10}$

Answer 3

$$x=\pm i$$ so $$a(\pm i)^3-a(\pm i)^2+b(\pm i)+4=0$$ so $$ai+a\pm bi+4=0$$ $$ai+a+bi+4=0\tag 1$$ or $$ai+a-bi+4=0\tag 2$$ now we will solve the first equation $$a+4=0\rightarrow a=-4$$ $$a-b=0\rightarrow b=-4$$ hence $$f(x)=-4x^3+4x^2-4x+4$$ at $x=4$ $$f(4)=-204=-4(51)$$

the second equation gives $$a=-4$$ $$b=4$$ hence $$f(x)=-4x^3+4x^2+4x+4$$ at $x=4$ $$f(4)=-176=-4(43)$$ as @marty said the problem is wrong

Answer 4

Let $a+b=c$.

Solving in pari/gp:

? lift(Mod(a*x^3-a*x^2+b*x+4,x^2+1))
%9 = (-a + b)*x + (a + 4)
?
? lift(Mod(a*x^3-a*x^2+b*x+4,x-4))
%10 = 48*a + (4*b + 4)
?
? polresultant(%9,a+b-c,a)
%11 = (-2*b + c)*x + (b + (-c - 4))
?
? polresultant(%10-51,a+b-c,a)
%12 = 44*b + (-48*c + 47)
?
? polresultant(%11,%12,b)
%13 = (52*c - 94)*x + (-4*c + 223)

Or in Wolfram.

I.e. $a+b=\dfrac{94x-223}{4(13x-1)}$.

If $x=4$ then $\begin{cases}4(b-a)+a+4=0\\48a+4b+4=51\end{cases} \Longrightarrow \begin{cases}a=1\\b=-1/4\end{cases} \Longrightarrow (a+b)\bigg|_{x=4}=\dfrac{3}{4}$.

Verifying: $(a+b)\bigg|_{x=4}=\dfrac{94\cdot 4-223}{4(13\cdot 4-1)}=\dfrac{3}{4}$.

By modulo $x^2+1$ remainder is $x(b-a)+a+4$ and quotient is $a(x-1)$:

$ax^3-ax^2+bx+4=\bigg(x(b-a)+a+4\bigg)+\bigg(a(x-1)\bigg)\cdot(x^2+1)$.

I think that from remainder$=0$ $\not\Rightarrow ax^3-ax^2+bx+4=a(x-1)(x^2+1)$, but only $ax^3-ax^2+bx+4 \equiv 0 \equiv x(b-a)+a+4 \equiv a(x-1)(x^2+1) \pmod {x^2+1}$. Thus is not correct to match the coefficients of $ax^3-ax^2+bx+4$ and $a(x-1)(x^2+1)$ whitout modulo.

Answer 5

This question is very interesting......

Only by this statement, the question can be solved

$f(x)= ax^3-ax^2+bx+4$ is a multiple of $x^2+1$;

Because the roots of the equation $x^2+1$ would be the roots of $f(x)$.

Therefore if $j,k$ are roots of the equation then $j+k=0$ and $j.k =1$. Let the roots of $f(x)$ be $j,k,l$.

Then,

$j+k+l = -\frac{-a}{a}$

$l=1$

now, $j.k.l= -\frac{4}{a}$

so, $a= -4$

now $j.k + l(k+j) = \frac{b}{-4}$

Therefore $b=-4$

Finally $a+b=-8$

But $f(4)= 4(-4^3+4^2-4+1)$
Therefore $f(4)= -204$

So the question may be incorrect.