Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively.

Question

So here is the Question :-

Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .

What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :- $$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$ $$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$ This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.

Any hints or explanations for this problem will be greatly appreciated !!

Answer 1

Hint

You have $$f(x)=(x^2-1)(ax+b)+(2x-5)$$ and $$f(x)=(x^2-4)(cx+d)+(-3x+4)$$ From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus $$\color{red}{c+d=\frac{4}{3}}$$

From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus $$\color{red}{2a+b=\frac{-1}{3}}$$ You also get $$f(0)=-b-5=-4d+4 \implies\color{red}{4d-b=9}$$

NOTE You can make a simple observation about the coefficients of $x^3$ and $x^2$ in both the expressions for $f(x)$ to conclude that $a=c$ and $b=d$. With that the first two equations can suffice.

Answer 2

Hint:

Let $$\dfrac{f(x)}{(x-1)(x+1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{x-2}+\dfrac D{x+2}$$

$$\implies f(x)=?$$

where $A,B,C,D$ which are arbitrary constants which can be found by putting $x=1,-1,2,-2$ one by one.

For example, put $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$

Answer 3

You have (for instance) that $f(x) = Q(x)(x+1)(x-1) + R(x)$, where $R(x) = 2x-5$.

Putting $x=1$ then $x=-1$ gives you $f(1) = - 3$ and $f(-1)=-7$.

Doing the same with $x=2, x=- 2$ allows you to work out two more points on the cubic.

Four distinct points allow you to completely characterise a cubic, you have a system of four linear equations to solve for the coefficients.

Answer 4

$f(x)$ leaves remainder $2x-5$ when divided by $x^2-1.$ Then, $$f(1)=2(1)-5=-3$$ $$f(-1)=2(-1)-5=-7$$

$f(x)$ leaves remainder $-3x+4$ when divided by $x^2-4.$ So we have, $$f(x)=(x^2-4)\,q(x)-3x+4$$ Since $f(x)$ is of degree $3$, the degree of $q(x)$ should be $1.$

Let $q(x)=ax+b.$ Then, $$f(x)=(x^2-4)(ax+b)-3x+4$$ $$f(x)=ax^3+bx^2-(4a+3)x-(4b-4)$$ Let $x=1$ then, $$f(1)=a(1)^3+b(1)^2-(4a+3)(1)-(4b-4)=-3$$ $$a+b=\dfrac{4}{3}\quad\quad\dots(1)$$ Let $x=-1$ then, $$f(-1)=a(-1)^3+b(-1)^2-(4a+3)(-1)-(4b-4)=-7$$ $$a-b=-\dfrac{14}{3}\quad\quad\dots(2)$$ Solving (1) and (2) we get, $$a=-\dfrac{5}{3}$$ $$b=3$$ Therefore, $$f(x)=-\dfrac{5}{3}x^3+3x^2+\dfrac{11}{3}x-8$$ Now, $$\underline{\underline{f(-3)=53}}$$

Answer 5

Working $\,\bmod (x^2-1)\,$ we have that $\,x^2 \equiv 1\,$, then:

$$ f(x) = ax^3+bx^2+cx+d\equiv ax+b+cx+d=(a+c)x+(b+d) $$

Comparing to the known remainder $\,2x-5\,$:

$$ \begin{cases} a+c=2 \\ b+d = -5 \tag{1} \end{cases} $$

Repeating the steps $\,\bmod (x^2-4)\,$ and remainder $\,-3x + 4\,$:

$$ f(x) \equiv 4ax+4b+cx+d=(4a+c)x+(4b+d) \;\;\implies\;\; \begin{cases} 4a+c=-3 \\ 4b+d = 4 \tag{2} \end{cases} $$

Solving the four equations $\,(1), (2)\,$ for $\,a,b,c,d\,$ gives the polynomial $\,f(x)\,$.