# Practice :The Remainder Theorem and The Factor Theorem

1.       It is given that $\displaystyle f(x)=x^3+ax^2+bx-48$. When $\displaystyle f(x)$ is divided by $\displaystyle x - 3$ the remainder is $\displaystyle 6$. Given that $\displaystyle f'(1)=0$ , find the value of $\displaystyle a$ and of $\displaystyle b$.

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$\displaystyle f(x)=x^3+ax^2+bx-48$

When $\displaystyle f(x)$ is divided by $\displaystyle x-3$, the remainder is $\displaystyle 6$.

$\displaystyle \begin{array}{l}\therefore \ \ f(3)=6\\\\\ \ \ \ {{(3)}^{3}}+a{{(3)}^{2}}+b(3)-48=6\\\\\ \ \ \ 3a+b=9\,\ \ -----(1)\\\\\ \ \ \ {f}'(x)=3{{x}^{2}}+2ax+b\\\\\ \ \ \ {f}'(1)=0\ \ \ \ [\text{given}]\\\\\ \ \ \ 3{{(1)}^{2}}+2a(1)+b=0\\\\\ \ \ \ 2a+b=-3\,\ \ ----(2)\end{array}$

By equation (1) - equation (2),

$\displaystyle \begin{array}{l}\ \ \ \ \ a=12\\\\\therefore \ \ \ 24+b=-3\, \Rightarrow b=-27\end{array}$

2.       If $\displaystyle f(x) = ax^2 + bx + c$ leaves remainders $\displaystyle 1 , 25 , 1$ on division by $\displaystyle x - 1 , x + 1 , x - 2$ respectively, show that $\displaystyle f(x)$ is a perfect square.

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$\displaystyle \begin{array}{l}\ \ \ f(x)=a{{x}^{2}}+bx+c\\\\\ \ \ f(1)=1\\\\\ \ \ a{{(1)}^{2}}+b(1)+c=1\\\\\ \ \ a+b+c=1\ \ \ \ ----(1)\\\\\ \ \ f(-1)=25\\\\\ \ \ a{{(-1)}^{2}}+b(-1)+c=1\\\\\ \ \ a-b+c=25\ ----(2)\\\\\ \ \ f(2)=1\\\\\ \ \ a{{(2)}^{2}}+b(2)+c=1\\\\\ \ \ 4a+2b+c=1\ \ ---(3)\\\\\ \ \ \text{By equation (1)}-\text{equation (2),}\\\\\ \ \ 2b=-24\Rightarrow b=-12\\\\\ \ \ \text{Substituting }b=-12\ \text{in equations (1)}\operatorname{and}\text{(3),}\\\\\ \ \ a-12+c=1\ \Rightarrow a+c=13\ \ \ \ \ --(4)\\\\\ \ \ 4a-24+c=1\ \Rightarrow 4a+c=25--(5)\\\\\ \ \ \text{By equation (5)}-\text{equation (4),}\\\\\ \ \ 3a=12\Rightarrow a=4\\\\\ \ \ \text{Substituting }a=4,b=-12\ \text{in equations (1), }\\\\\ \ \ 4-12+c=1\ \Rightarrow c=9\\\\\therefore f(x)=4{{x}^{2}}-12x+9={{(2x-3)}^{2}}\\\\\therefore f(x)\ \text{is a perfect square}\text{.}\end{array}$

3.       It is given that $\displaystyle x-2$ is a factor of $\displaystyle f(x)=x^3+kx^2-8x-8$ where $\displaystyle k$ is an integer.

(i) Find the value of the integer $\displaystyle k$.

(ii) Using your of $\displaystyle k$, find the non-integer roots of the equation $\displaystyle f(x)=0$ in the form $\displaystyle a\pm \sqrt{b}$, where $\displaystyle a$ and $\displaystyle b$ are integers.

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$\displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}+k{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Since}\ x-2\ \text{is a factor of }f(x),\\\\\ \ \ \ f(2)=0\\\\\therefore \ \ {{(2)}^{3}}+k{{(2)}^{2}}-8(2)-8=0\\\\\therefore \ \ 8+4k-16-8=0\\\\\therefore \ \ k=4\\\\\therefore \ \ f(x)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Let}\ f(x)=(x-2)({{x}^{2}}+ax+b)\\\\\therefore \ \ (x-2)({{x}^{2}}+ax+b)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ {{x}^{3}}+(a-2){{x}^{2}}+(b-2a)x-2b={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ a-2=4\Rightarrow a=6\\\\\ \ \ \ 2b=8\Rightarrow b=4\\\\\therefore \ \ f(x)=(x-2)({{x}^{2}}+6x+4)\\\\\ \ \ \ \text{When }f(x)=0,\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+4)=0\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+9-5)=0\\\\\ \ \ \ (x-2)\left[ {{{{(x+3)}}^{2}}-5} \right]=0\\\\\ \ \ \ x=2\ \text{or }{{(x+3)}^{2}}=5\\\\\therefore \ \ x=2\ \text{or }x=-3\pm \sqrt{5}\\\\\therefore \ \ \text{The non-integer roots of }f(x)=0\ \text{are }-3\pm \sqrt{5}.\end{array}$

4.       A function f is such that $\displaystyle f(x)=4x^3+4x^2+ax+b$. It is given that $\displaystyle 2x−1$ is a factor of both $\displaystyle f(x)$ and $\displaystyle f'(x)$.

(i) Find the value of $\displaystyle a$ and $\displaystyle b$.

(ii) Hence find the remainder when $\displaystyle f(x)$ is divided by $\displaystyle x+3$.

(iii) Express $\displaystyle f(x)$ in the form $\displaystyle f(x)=(2x−1)(px^2+qx+r)$, where $\displaystyle p, q$ and $\displaystyle r$ are integers to be found.

(iv) find the values of $\displaystyle x$ for which $\displaystyle f(x) = 0$.

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$\displaystyle \begin{array}{l}\ \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}+ax+b\\\\\therefore \ \ {f}'(x)=12{{x}^{2}}+8x+a\\\\\ \ \ \ 2x-1\ \text{is a factor of both}\ f(x)\ \text{and}\ {f}'(x).\end{array}$

$\displaystyle \therefore \ \ f\left( {\frac{1}{2}} \right)=0$

$\displaystyle \therefore \ \ 4{{\left( {\frac{1}{2}} \right)}^{3}}+4{{\left( {\frac{1}{2}} \right)}^{2}}+a\left( {\frac{1}{2}} \right)+b=0$

$\displaystyle \ \ \ \ \frac{1}{2}+1+\frac{a}{2}+b=0$

$\displaystyle \therefore \ \ \ a+2b=-3$

$\displaystyle \ \ \ \ \ \text{And}\ {f}'\left( {\frac{1}{2}} \right)=0$

$\displaystyle \therefore \ \ \ 12{{\left( {\frac{1}{2}} \right)}^{2}}+8\left( {\frac{1}{2}} \right)+a=0$

$\displaystyle \begin{array}{l}\therefore \ \ \ a=-7\\\\\therefore \ \ \ -7+2b=-3\ \Rightarrow b=2\\\\\therefore \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \text{When}\ f(x)\ \text{is divided by}x+3,\\\\\ \ \ \ \ \text{the remainder is}\ f(-3).\\\\\therefore \ \ \ f(-3)=4{{(-3)}^{3}}+4{{(-3)}^{2}}-7(-3)+2=-49\\\\\ \ \ \ \ \text{When}\ f(x)=(2x-1)(p{{x}^{2}}+qx+r),\\\\\ \ \ \ \ \ (2x-1)(p{{x}^{2}}+qx+r)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \ 2p{{x}^{3}}+(2q-p){{x}^{2}}+(2r-q)x-r=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\therefore \ \ \ \ r=-2\\\\\ \ \ \ \ \ 2r-q=-7\Rightarrow -4-q=-7\Rightarrow q=3\\\\\ \ \ \ \ \ 2q-p=4\Rightarrow 6-p=4\Rightarrow p=2\\\\\therefore \ \ \ \ f(x)=(2x-1)(2{{x}^{2}}+3x-2)\\\\\ \ \ \ \ \text{When}\ f(x)=0,(2x-1)(2{{x}^{2}}+3x-2)=0\\\\\therefore \ \ \ \ (2x-1)(2x-1)(x+2)=0\end{array}$

$\displaystyle \therefore \ \ \ \ x=\frac{1}{2}\ \text{or}\ x=-2$

5.       The remainder when the expression $\displaystyle x^3+9x^2+bx+c$ is divided by $\displaystyle x-2$ is twice the remainder when the expression is divided by $\displaystyle x-1$. Show that $\displaystyle c = 24$. Given that $\displaystyle x+8$ is a factor of $\displaystyle x^3+9x^2+bx+24$, show that the equation $\displaystyle x^3+9x^2+bx+c=0$ has only one real root.

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$\displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)={{x}^{3}}+9{{x}^{2}}+bx+c\\\\\ \ \ \ \text{By the problem},\\\\\ \ \ \ f(2)=2f(1)\\\\\ \ \ \ {{2}^{3}}+9\cdot {{2}^{2}}+2b+c=2\left( {{{1}^{3}}+9\cdot {{1}^{2}}+b+c} \right)\\\\\ \ \ \ 8+36+2b+c=2+18+2b+2c\\\\\therefore \ \ c=24\\\\\ \ \ \ \text{Since}\ x+8\ \text{is a factor of}\ {{x}^{3}}+9{{x}^{2}}+bx+24,\\\\\ \ \ \ f(-8)=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\therefore \ \ b=11\\\\\therefore \ f(x)={{x}^{3}}+9{{x}^{2}}+11x+24\end{array}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}+\ \ x+\ \ 3\\x+8\ \ \overline{\left){\begin{array}{l}{{x}^{3}}+9{{x}^{2}}+11x+24\\\underline{{{{x}^{3}}+8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ {{x}^{2}}+11x\\\ \ \ \ \ \ \ \ \underline{{{{x}^{2}}+\ \ 8x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x+24\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{3x+24}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$\displaystyle \therefore \ \ f(x)=(x+8)({{x}^{2}}+x+3)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{x}^{2}}+2\left( {\frac{1}{2}} \right)x+\frac{1}{4}+\frac{{11}}{4}} \right]$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]$

$\displaystyle \ \ \ \ \text{When}\ f(x)=0,(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]=0$

$\displaystyle \therefore \ \ x=-8\ \text{or}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}$

$\displaystyle \ \ \ \ \text{Since}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}\ge 0,{{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}\ \text{is impossible}\text{.}$

$\displaystyle \therefore \ \ \text{There is no other real solution for }{{x}^{3}}+9{{x}^{2}}+11x+24=0.$

6.       Given that the remainder when $\displaystyle f(x) = x^3 - x^2 + ax$ is divided by $\displaystyle x + a$, where $\displaystyle a > 0$, is twice the remainder when $\displaystyle f(x)$ is divided by $\displaystyle x - 2a$, find the value of $\displaystyle a$. Find also the remainder when $\displaystyle f(x)$ is divided by $\displaystyle x – 2$.

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$\displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}-{{x}^{2}}+ax\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ f(-a)=2f(2a)\\\\\ \ \ \ -{{a}^{3}}-{{a}^{2}}-{{a}^{2}}=2\left( {8{{a}^{3}}-4{{a}^{2}}+2{{a}^{2}}} \right)\\\\\ \ \ \ 17{{a}^{3}}-2{{a}^{2}}=0\\\\\ \ \ \ {{a}^{2}}(17a-2)=0\end{array}$

$\displaystyle \ \ \ \ \text{Since}\ a>0,17a-2=0\Rightarrow a=\frac{2}{{17}}$

$\displaystyle \therefore \ \ f(x)={{x}^{3}}-{{x}^{2}}+\frac{2}{{17}}x$

$\displaystyle \begin{array}{l}\ \ \ \ \text{When}\ f(x)\ \text{is divided by}\ x-2,\\\ \ \ \ \text{the remainder is}\ f(2).\end{array}$

$\displaystyle \therefore \ \ f(2)={{(2)}^{3}}-{{(2)}^{2}}+\frac{2}{{17}}(2)=\frac{{72}}{{17}}$

7.       Given that $\displaystyle f(x) = x^{2n} - ( p + 1)x^2 + p$, where $\displaystyle n$ and $\displaystyle p$ are positive intergers. Show that $\displaystyle x - 1$ is a factor of $\displaystyle f(x)$ for all values of $\displaystyle p$. When $\displaystyle p = 4$, find the value of $\displaystyle n$ for which $\displaystyle x - 2$ is a factor of $\displaystyle f(x)$ and, for this case, hence factorise $\displaystyle f(x)$ completely.

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$\displaystyle \begin{array}{l}\ \ \ \ \ f(x)={{x}^{{2n}}}-(p+1){{x}^{2}}+p\ \text{where}\ n,p\in {{J}^{+}}\\\\\ \ \ \ \ f(1)={{(1)}^{{2n}}}-(p+1){{(1)}^{2}}+p\\\\\ \ \ \ \ \text{Since}\ n,p\in {{J}^{+}},{{(1)}^{{2n}}}=1\ \text{and}-(p+1)=-p-1.\\\\\therefore \ \ \ f(1)=1-p-1+p=0\\\\\therefore \ \ \ x-1\ \text{is a factor of }f(x)\ \text{for all values of}\ p.\\\\\ \ \ \ \ \text{When}\ p=4,f(x)={{x}^{{2n}}}-5{{x}^{2}}+4.\\\\\ \ \ \ \ x-2\ \text{is a factor of }f(x).\\\\\therefore \ \ \ f(2)=0\\\\\ \ \ \ \ {{2}^{{2n}}}-5\cdot {{2}^{2}}+4=0\\\\\therefore \ \ \ {{2}^{{2n}}}=16\Rightarrow n=2\end{array}$

8.       The polynomial $\displaystyle f(x)$ is given by $\displaystyle f(x) = ax^3 + 11x^2 + cx - 60$ , where $\displaystyle a$ and $\displaystyle c$ are constants. If the roots of $\displaystyle f(x) = 0$ are $\displaystyle 2 , - 3$ and $\displaystyle k$, find the values of $\displaystyle a , c$ and $\displaystyle k$.

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$\displaystyle \begin{array}{l}\ \ \ \ \ f(x)=a{{x}^{3}}+11{{x}^{2}}+cx-60\\\\\ \ \ \ \ \text{Since the roots of}\ f(x)=0\ \text{are}\ 2,-3\ \text{and}\ k,\\\\\ \ \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60=(x-2)(x+3)(x-k)\\\\\therefore \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60={{x}^{3}}+(1-k){{x}^{2}}-(6+k)x+6k\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ a=1,\\\\\ \ \ \ \ 6k=-60\Rightarrow k=-10\\\\\ \ \ \ \ c=-(6+k)=-(6-10)=4\end{array}$

9.       Given that $\displaystyle 4x^4 - 9a^2x^2 + 2(a^2 - 7) x - 18$ is exactly divisible by $\displaystyle 2x - 3a$, show that $\displaystyle a^3 - 7a - 6 = 0$ and hence find the possible values of $\displaystyle a$.

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$\displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)=4{{x}^{4}}-9{{a}^{2}}{{x}^{2}}+2({{a}^{2}}-7)x-18\\\\\ \ \ \ \ \text{Since }f(x)\ \text{is exactly divisible by}\ 2x-3a,\end{array}$

$\displaystyle \ \ \ \ \ f\left( {\frac{{3a}}{2}} \right)=0.$

$\displaystyle \therefore \ \ \ 4{{\left( {\frac{{3a}}{2}} \right)}^{4}}-9{{a}^{2}}{{\left( {\frac{{3a}}{2}} \right)}^{2}}+2({{a}^{2}}-7)\left( {\frac{{3a}}{2}} \right)-18=0$

$\displaystyle \ \ \ \ \ \ 4\left( {\frac{{81{{a}^{4}}}}{{16}}} \right)-9{{a}^{2}}\left( {\frac{{9{{a}^{2}}}}{4}} \right)+3{{a}^{3}}-21a-18=0$

$\displaystyle \begin{array}{l}\therefore \ \ \ \ {{a}^{3}}-7a-6=0\\\\\ \ \ \ \ \text{Let }g(a)={{a}^{3}}-7a-6\\\\\ \ \ \ \ g(-1)=-1+7-6=0\\\\\therefore \ \ \ \ a+1\ \text{is a factor of }g(a).\\\\\therefore \ \ \ \ \text{Let }g(a)=(a+1)({{a}^{2}}+pa+q),\\\ \ \ \ \ \ \text{where }p\ \text{and }q\ \text{are any real constants}\text{.}\\\\\therefore \ \ \ g(a)={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\therefore \ \ \ {{a}^{3}}-7a-6={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ p+1=0\Rightarrow p=-1\ \ q=-6\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)({{a}^{2}}-a-6)\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)(a+2)(a-3)\\\\\ \ \ \ \text{When }{{a}^{3}}-7a-6=0,\\\\\ \ \ \ (a+1)(a+2)(a-3)=0\\\\\therefore \ \ a=-1\ \text{or }a=-2\ \text{or }a=3.\end{array}$

10.      Solve the equation $\displaystyle 4x^3 + 3x^2 -16x = 12$. Hence find the value of $\displaystyle x$ such that $\displaystyle 4e^{3x} + 3e^{2x} - 16e^x = 12$.

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$\displaystyle \begin{array}{l}\ \ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x=12\\\\\therefore \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\ \ \ \ \ \text{Let}\ f(x)=4{{x}^{3}}+3{{x}^{2}}-16x-12\\\\\ \ \ \ \ f(2)=4{{(2)}^{3}}+3{{(2)}^{2}}-16(2)-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =32+12-32-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\\therefore \ \ \ (x-2)\ \text{is a factor of }f(x).\end{array}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{x}^{2}}+11x+\ \ 6\\\ \ \ \ \ x-2\overline{\left){\begin{array}{l}4{{x}^{3}}+3{{x}^{2}}-16x-12\\\underline{{4{{x}^{3}}-8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ 11{{x}^{2}}-16x\\\ \ \ \ \ \ \ \ \ \ \underline{{11{{x}^{2}}-22x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x-12\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{6x-12}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$\displaystyle \begin{array}{l}\therefore \ \ f(x)=(x-2)\left( {4{{x}^{2}}+11x+\ \ 6} \right)\\\\\therefore \ \ f(x)=(x-2)(x+2)(4x+3)\\\\\ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\therefore \ \ (x-2)(x+2)(4x+3)=0\end{array}$

$\displaystyle \therefore \ \ x=2\ \text{or}\ x=-2\ \text{or}\ x=-\frac{3}{4}$

$\displaystyle \begin{array}{l}\ \ \ \ 4{{e}^{{3x}}}+3{{e}^{{2x}}}-16{{e}^{x}}=12\\\\\ \ \ \ 4{{({{e}^{x}})}^{3}}+3{{({{e}^{x}})}^{2}}-16{{e}^{x}}-12=0\end{array}$

$\displaystyle \therefore \ \ {{e}^{x}}=2\ \text{or}\ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}$

$\displaystyle \ \ \ \ \text{Since}\ {{e}^{x}}>0\ \text{for all }x\in R,$

$\displaystyle \ \ \ \ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}\ \text{is impossible}$

$\displaystyle \therefore \ \ {{e}^{x}}=2\Rightarrow x=\ln 2\$

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