Practice Problems : The Remainder and Factor Theorem

1.        What number should be added to $ \displaystyle 2x^3 - 3x^2 - 8x$ so that the resulting polynomial leaves the remainder $ \displaystyle 10$ when divided by $ \displaystyle 2x + 1$?

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Let the number to be added be $ \displaystyle k$ and the resulting polynomial be $ \displaystyle f(x)$. $ \displaystyle \therefore \ f(x)=2{{x}^{3}}-3{{x}^{2}}-8x+k$ When $ \displaystyle f(x)$ is divided by $ \displaystyle 2x+1$, the remainder is 10. $ \displaystyle \begin{array}{l}\therefore \ f\left( {-\displaystyle \frac{1}{2}} \right)=10\\\\\ \ 2{{\left( {-\displaystyle \frac{1}{2}} \right)}^{3}}-3{{\left( {-\displaystyle \frac{1}{2}} \right)}^{2}}-8\left( {-\displaystyle \frac{1}{2}} \right)+k=10\\\\\therefore \ -\displaystyle \frac{1}{4}-\displaystyle \frac{3}{4}+4+k=10\\\\\therefore \ k=7\end{array}$ Hence, the number to be added is 10.

2.        What number should be subtracted from $ \displaystyle 6x^3 + 7x^2 - 9x+12$ so that $ \displaystyle 3x - 1$ is the factor of the resulting polynomial?

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Let the number to be subtracted be $ \displaystyle k$ and the resulting polynomial be $ \displaystyle f(x)$. $ \displaystyle \therefore \ f(x)=6x^3 + 7x^2 - 9x+12-k$ Since $ \displaystyle 3x - 1$ is the factor of $ \displaystyle f(x)$, $ \displaystyle \begin{array}{l}\ \ f\left( {\displaystyle \frac{1}{3}} \right)=0\\\\\ \ 6{{\left( {\displaystyle \frac{1}{3}} \right)}^{3}}+7{{\left( {\displaystyle \frac{1}{3}} \right)}^{2}}-9\left( {\displaystyle \frac{1}{3}} \right)+12-k=0\\\\\therefore \ \displaystyle \frac{2}{9}+\displaystyle \frac{7}{9}-3+12-k=0\\\\\therefore \ k=10\end{array}$ Hence, the number to be subtracted is 10.

3.        When divided by $ \displaystyle x - 3$ the polynomials $ \displaystyle x^3 - px^2 + x + 6$ and $ \displaystyle 2x^3 - x^2 - (p + 3) x - 6$ leave the same remainder. Find the value of $ \displaystyle p$.

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Let $ \displaystyle f(x)={{x}^{3}}-p{{x}^{2}}+x+6$ and $ \displaystyle g(x)=2x^3 - x^2 - (p + 3) x - 6$ $ \displaystyle f(x)$ and $ \displaystyle g(x)$ leave the same remainder when divided by $ \displaystyle x - 3$, $ \displaystyle \begin{array}{l}\therefore \ f(3)=g(3)\\\\\ \ \ {{(3)}^{3}}-p{{(3)}^{2}}+(3)+6=2{{(3)}^{3}}-{{(3)}^{2}}-(p+3)(3)-6\\\\\ \ \ 27-9p+9=54-9-3p-9-6\\\ \ \\\therefore \ 36-9p=30-3p\ \\\\\therefore \ p=1\ \end{array}$

4.       Using remainder theorem, find the value of $ \displaystyle a$ if the division of $ \displaystyle x^3 + 5x^2 - ax + 6$ by $ \displaystyle x -1$ leaves the remainder $ \displaystyle 2a$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+5{{x}^{2}}-ax+6\\\\\ \ \ \text{When}\ f(x)\ \text{is divided by }x-1,\ \\\\\ \ \ \text{the remainder}\ =2a\\\\\therefore \ f(1)=2a\\\\\ \ \ {{1}^{3}}+5{{(1)}^{2}}-a(1)+6=2a\\\\\ \ \ 1+5-a+6=2a\\\\\therefore \ 3a=12\\\\\ \ \ a=4\ \ \ \end{array}$

5.        Find the value of the constants $ \displaystyle a$ and $ \displaystyle b$, if $ \displaystyle x - 2$ and $ \displaystyle x + 3$ are both factors of the expression $ \displaystyle x^3 + ax^2 + bx - 12$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+a{{x}^{2}}+bx-12\\\\\ \ \ (x-2)\ \text{and}\ (x+3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(2)=0\\\\\ \ \ {{(2)}^{3}}+a{{(2)}^{2}}+b(2)-12=0\\\\\therefore \ \ 8+4a+2b-12=0\\\\\therefore \ 2a+b=2---(1)\\\\\ \ \text{Again}\ f(-3)=0\\\\\ \ \ {{(-3)}^{3}}+a{{(-3)}^{2}}+b(-3)-12=0\\\\\therefore \ \ -27+9a-3b-12=0\\\\\therefore \ \ 3a-b=13---(2)\\\\\ \ \ (1)+(2)\Rightarrow 5a=15\\\\\therefore \ \ a=3\\\\\therefore \ 2\left( 3 \right)+b=2\\\\\therefore \ b=-4\end{array}$

6.        If $ \displaystyle x + 2$ and $ \displaystyle x - 3$ are factors of $ \displaystyle x^3 + ax + b$, find the values of $ \displaystyle a$ and $ \displaystyle b$.
With these values of $ \displaystyle a$ and $ \displaystyle b$, factorise the given expression.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+ax+b\\\\\ \ \ (x+2)\ \text{and}\ (x-3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(-2)=0\\\\\ \ \ {{(-2)}^{3}}+a(-2)+b=0\\\\\therefore \ \ -8-2a+b=0\\\\\therefore \ -2a+b=8---(1)\\\\\ \ \text{Again}\ f(3)=0\\\\\ \ \ {{(3)}^{3}}+a(3)+b=0\\\\\therefore \ \ 27+3a+b=0\\\\\therefore \ \ 3a+b=-27---(2)\\\\\ \ \ (2)-(1)\Rightarrow 5a=-35\\\\\therefore \ \ a=-7\\\\\therefore \ \ -2\left( {-7} \right)+b=8\\\\\therefore \ \ b=-6\\\\\therefore \ \ f(x)={{x}^{3}}-7x-6\\\\\ \ \ \ \text{Let}\ {{x}^{3}}-7x-6=(x+2)(x-3)(x+k)\\\\\therefore \ \ {{x}^{3}}-7x-6={{x}^{3}}+\left( {k-1} \right){{x}^{2}}-\left( {k+6} \right)x-6k\\\\\therefore \ k-1=0\Rightarrow k=1\\\\\therefore \ \ f(x)=(x+2)(x-3)(x+1)\\\ \ \end{array}$

7.        Given that $ \displaystyle x - 2$ is a factor of the expression $ \displaystyle x^3 + ax^2 + bx + 6$.
When this expression is divided by $ \displaystyle x - 3$, it leaves the remainder $3$. Find the values of $ \displaystyle a$ and $ \displaystyle b$.

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Let $f(x)=x^{3}+a x^{2}+b x+6\\\\ $ $x-2$ is a factor of $f(x)$. $\begin{aligned} &\\ &\therefore\ f(2)=0\\\\ &2^{3}+a(2)^{2}+2 b+6=0\\\\ &2 a+b=-7 \ldots(1)\\\\ \end{aligned}$ When $f(x)$ is divided by $x-3$, the remainder is $3$. $\begin{aligned} &\\ &\therefore\ f(3) =3 \\\\ &3^{3}+a(3)^{2}+3 b+6 =3 \\\\ &3 a+b =-10 \cdots(2)\\\\ \end{aligned}$ By equation $(2)$ - equation $(1)$, $\begin{aligned} &\\ a=-3\\\\ \end{aligned}$ Substituting $a=-3$ in equation $(1)$, $\begin{aligned} &\\ &2(-3)+b=-7\\\\ &\therefore\ b=-1 \end{aligned}$

8.        If $ \displaystyle x - 2$ is a factor of the expression $ \displaystyle 2x^3 + ax^2 + bx - 14$ and when the expression is divided by $ \displaystyle x - 3$,
it leaves a remainder $ \displaystyle 52$, find the values of $ \displaystyle a$ and $ \displaystyle b$.

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Let $f(x)=2x^{3}+a x^{2}+b x-14\\\\ $ $x-2$ is a factor of $f(x)$. $\begin{aligned} &\\ &\therefore\ f(2)=0\\\\ &2(2)^{3}+a (2)^{2}+b (2)-14=0\\\\ &2 a+b=-1 \ldots(1)\\\\ \end{aligned}$ When $f(x)$ is divided by $x-3$, the remainder is $52$. $\begin{aligned} &\\ &\therefore\ f(3) =52 \\\\ &2(3)^{3}+a (3)^{2}+b (3)-14 =52 \\\\ &3 a+b =4 \cdots(2)\\\\ \end{aligned}$ By equation $(2)$ - equation $(1)$, $\begin{aligned} &\\ a=5\\\\ \end{aligned}$ Substituting $a=5$ in equation $(1)$, $\begin{aligned} &\\ &2(5)+b=-1\\\\ &\therefore\ b=-11 \end{aligned}$

9.        If $ ax^3 + 3x^2 + bx - 3$ has a factor $2x + 3$ and leaves remainder $-3$
when divided by $x + 2$, find the values of $ a$ and $ \displaystyle b$. With these values of $a$ and $ b$, factorise the given expression.

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Let $f(x)=ax^3 + 3x^2 + bx - 3\\\\ $ $2x+3$ is a factor of $f(x)$. $\begin{aligned} &\\ &\therefore\ f\left(-\dfrac{3}{2}\right)=0\\\\ &a\left(-\dfrac{3}{2}\right)^3 + 3\left(-\dfrac{3}{2}\right)^2 + b\left(-\dfrac{3}{2}\right) - 3=0\\\\ &-\frac{27}{8}a-\frac{3}{2}b + \frac{15}{4}=0\\\\ &9 a+4b=10 \ldots(1)\\\\ \end{aligned}$ When $f(x)$ is divided by $x+2$, the remainder is $-3$. $\begin{aligned} &\\ &\therefore\ f(-2) =-3 \\\\ &a(-2)^3 + 3(-2)^2 + b(-2) - 3 =-3 \\\\ &4 a+b =6 \\\\ &b =6-4a \cdots(2)\\\\ \end{aligned}$ Substituting $b =6-4a$ in equation $(1)$, $\begin{aligned} &\\ 9 a+4(6-4a)=10\\\\ &a=2\\\\ \end{aligned}$ Substituting $a=2$ in equation $(2)$, $\begin{aligned} &\\ &b =6-4(2)\\\\ &\therefore\ b=-2\\\\ &\therefore\ f(x)=2x^3 + 3x^2 -2x - 3\\\\ \end{aligned}$ Since $f(x)$ is a cubic polynomial with factor $2x+3$ and leading coefficient $2$, assume that $f(x) = (2x+3)(x^2 + px + q)$. $\begin{aligned} &\\ &2x^3 + 3x^2 -2x - 3=(2x+3)(x^2 + px + q)\\\\ &\therefore\ 2x^3 + 3x^2 -2x - 3=2x^3 + (2p+3)x^2 + (3p+2q)x + 3q\\\\ &\text{Equating the respective terms}\\\\ & 3q=-3\\\\ &\therefore\ q=-1\\\\ & 2p+3 = 3\\\\ &\therefore\ p=0\\\\ &\therefore\ f(x) = (2x+3)(x^2 -1)\\\\ &\therefore\ f(x) = (2x+3)(x +1)(x-1) \end{aligned}$

10.      Given $ \displaystyle f (x) = ax^2 + bx + 2$ and $ \displaystyle g (x) = bx^2 + ax + 1$. If $ \displaystyle x - 2$ is a factor of $ \displaystyle f (x)$
but leaves the remainder $ \displaystyle -15$ when it divides $ \displaystyle g (x)$, find the values of $ \displaystyle a$ and $ \displaystyle b$. With these values of $ \displaystyle a$ and $ \displaystyle b$,
factorise the expression $ \displaystyle f (x) + g (x) + 4x^2 + 7x$.

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$\begin{aligned} &f(x)=a x^{2}+b x+2\\\\ &x-2 \text{ is a factor of } f(x)\\\\ &\therefore\ a(2)^{2}+2 b+2=0\\\\ &2 a+b=-1 \ldots(1)\\\\ &g(x)=b x^{2}+a x+1\\\\ \end{aligned}$ When g(x) is divided by $x-2$, the remainder is $-15$. $\begin{aligned} &\\ \therefore g(2)&=-15 \\\\ b(2)^{2}+2 a+1 &=-15 \\\\ a+2 b &=-8 \ldots(2) \\\\ (1)+(2) \Rightarrow 3 a+3 b &=-9 \\\\ a+b &=-3 \ldots(3) \\\\ (1)-(2) \Rightarrow a-b &=7 \ldots(4) \\\\ (3)+(4) \Rightarrow 2 a &=4 \\\\ a &=2 \\\\ (3)-(4) \Rightarrow 2 b &=-10 \\\\ b &=-5\\\\ \end{aligned}$ $\begin{aligned} \text { Let } h(x) &=f(x)+g(x)+4 x^{2}+7 x \\\\ &=2 x^{2}-5 x+2-5 x^{2}+2 x+1+4 x^{2}+7 x \\\\ &=x^{2}-4 x+3 \\\\ &=(x-1)(x-3) \end{aligned}$

11.     When $ \displaystyle x^3 - 2x^2 + px - q$ is divided by $ \displaystyle x^2 - 2x - 3$, the remainder is $ \displaystyle x - 6$, What are the values of $ \displaystyle p$ and $ \displaystyle q$
respectively ?

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Let $f(x)=x^{3}-2 x^{2}+p x-q\\\\ $. When $f(x)$ is divided by $x^{2}-2 x-3$, the remainder is $x-6\\\\ $. Let $Q(x)$ be the quotient when $f(x)$ is divided by $x^{2}-2 x-3$. $\begin{aligned} & \\ & \therefore\ f(x)=Q(x)\left(x^{2}-2 x-3\right)+x-6 \\\\ & x^{3}-2 x^{2}+p x-q=Q(x)(x+1)(x-3)+x-6\\\\ & \text{When } x=-1,\\\\ &(-1)^{3}-2(-1)^{2}-p-q=-7 \\\\ & p+q=4 \ldots(1)\\\\ & \text{When } x=3,\\\\ &(3)^{3}-2(3)^{2}+p(3)-q=-3 \\\\ & 3 p-q=-12 \ldots(2)\\\\ & \text{By equation} (1) + \text{ equation } (2)\\\\ & 4 p =-8 \\\\ & p =-2\\\\ &\text{Substituting } p=-2 \text{ in equation } (1),\\\\ & -2+q=4 \\\\ & q=6 \end{aligned}$

12.      If $ \displaystyle x + k$ is a common factor of $ \displaystyle x^2 + px + q$ and $ \displaystyle x^2 + lx + m$, Find the value of $ \displaystyle k$
in terms of $ \displaystyle p, q, l$ and $ \displaystyle m$.

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$\begin{aligned} \text { Let } f(x) &=x^{2}+p x+q \\\\ g(x) &=x^{2}+l x+m \\\\ \end{aligned}$ $x+k$ is a common factor of $f(x)$ and $g(x)$. $\begin{aligned} & \\ & \therefore \quad f(-k)=0 \\\\ & (-k)^{2}+p(-k)+q=0 \\\\ & k^{2}-p k+q=0 \ldots(1) \\\\ & g(-k)=0 \\\\ &(-k)^{2}+L(-k)+m \\\\ & k^{2}-l k+m=0 \ldots(2)\\\\ \end{aligned}$ By equation (1) - equation (2), $\begin{aligned} &\\ & -p k+l k +q-m=0 \\\\ & (l-p) k =m-q \\\\ & k =\dfrac{m-q}{l-p} \end{aligned}$

13.      When a polynomial $ \displaystyle f(x)$ is divided by $ \displaystyle x - 3$ and $ \displaystyle x + 6$, the respective remainders are $ \displaystyle 7$ and $ \displaystyle 22$. What is the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle (x - 3) (x + 6)$ ?

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Let $f(x)$ be a polynomial.By the problem, $\begin{aligned} &\\ f(3)&=7 \\\\ f(-6)&=22\\\\ \end{aligned}$ Let $Q(x)$ be the quotiont and $a x+b$ be the remainder When $f(x)$ is divided by $(x-3)(x+6)$ $\begin{aligned} &\\ \therefore f(x) &=Q(x)(x-3)(x+b)+a x+b \\\\ f(3) &=3 a+b=7 \ldots(1) \\\\ f(-6) &=-6 a+b=22 \ldots(2)\\\\ \end{aligned}$ By equation $(1)$ - equation $(2)$, $\begin{aligned} &\\ 9 a &=-15 \\\\ a &=-\dfrac{5}{3}\\\\ \end{aligned}$ Substituting $a=-\dfrac{5}{3}$ in equation $(1)$, $\begin{aligned} &\\ 3\left(-\frac{5}{3}\right)+b=7 \\\\ b=12\\\\ \end{aligned}$ $\therefore$ When $f(x)$ is divided by $(x-3)(x+6)$, the remainder is $-\dfrac{5}{3} x+12$. $ \displaystyle (x-3)(x+6)$ သည္ polynomial of second degree ျဖစ္ပါသည္။ $ \displaystyle f(x)$ ကို $ \displaystyle (x-3)(x+6)$ ႏွင့္ စားေသာ အႂကြင္းသည္ စားကိန္းေအာက္ တစ္ထပ္ေလ်ာ့ပါမည္။။ ထို႔ေၾကာင့္ အႂကြင္းသည္ $ \displaystyle ax + b$ ပံုစံျဖစ္ပါမည္။

14.      Given that $ \displaystyle f (x) = x^3 + ax^2 + bx + c$. If $ \displaystyle f (1) = f(2) = 0$ and $ \displaystyle f(4) = f(0)$, find $ \displaystyle a, b$ and $ \displaystyle c$.

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$\begin{aligned} &f(x)=x^{3}+a x^{2}+b x+c \\\\ &f(1)=0 \\\\ &1^{3}+a(1)^{2}+b(1)+c=0 \\\\ &a+b+c=-1 \cdots(1) \\\\ &f(2)=0 \\\\ &2^{3}+a(2)^{2}+b(2)+c=0 \\\\ &4 a+2 b+c=-8 \ldots(2)\\\\ &f(4)=f(0) \\\\ &4^{3}+a(4)^{2}+b(4)+c=c \\\\ &4 a+b=-16 \ldots(3)\\\\ \end{aligned}$ By equation $(2)$ - equation $(1)$, $\begin{aligned} &\\ 3 a+b&=-7 \ldots(4)\\\\ \end{aligned}$ By equation $(3)$ - equation $(4)$, $\begin{aligned} &\\ a=-9\\\\ \end{aligned}$ Substituting $a=-9$ in equation $(4)$, $\begin{aligned} &\\ 3(-9)+b &=-7 \\\\ b &=20\\\\ \end{aligned}$ Substituting $a=-9$ and $b=20$ in equation $(1)$, $\begin{aligned} &\\ -9+20+c &=-1 \\\\ c &=-12 \end{aligned}$

15.      If $ \displaystyle x – 1$ is a factor of $ \displaystyle Ax^3 + Bx^2 - 36x + 22$ and $ \displaystyle 2^B = 64^A$, find $ \displaystyle A$ and $ \displaystyle B$.

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$\begin{aligned} &\text { Let } f(x)=A x^{3}+B x^{2}-36 x+22. \\\\ &x-1 \text { is a factor of } f(x). \\\\ &\therefore\ A+B-36+22=0 \\\\ &A+B=14 \ldots (1) \\\\ &2^{B}=64^{A}(\text { given }) \\\\ &2^{B}=2^{6 A} \\\\ &\therefore B=6 A \ldots (2)\\\\ \end{aligned}$ Substituting $B=6 A$ in equation $(1)$, $\begin{aligned} &\\ A+6 A &=14 \\\\ 7 A &=14 \\\\ A &=2\\\\ \end{aligned}$ Substituting $A=2$ in equation $(2)$. $\begin{aligned} &\\ B & =6(2)\\\\ &=12\\\\ \end{aligned}$

16.      When k is subtracted from $ \displaystyle 27x^3 - 9x^2 - 6x - 5$, it is exactly divisible by $3x - 1$, find $ \displaystyle k$.

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Let $f(x)=27 x^{3}-9 x^{2}-6 x-5-k\\\\ $ $f(x)$ is divisible by $3 x-1$, $\begin{aligned} &\\ & \therefore\ f\left(\dfrac{1}{3}\right)=0 \\\\ & 27\left(\dfrac{1}{3}\right)^{3}-9\left(\dfrac{1}{3}\right)^{2}-6\left(\dfrac{1}{3}\right)-5-k=0 \\\\ & 1-1-1-2-5-k=0 \\\\ & k=-8 \end{aligned}$

17.      Given that $f(x)=4x^3-4kx^2-x+k,\ \ $ $ g(x)=3x^2+(1-3k)x-k$ and $h(x)=f(x)+g(x)$. If $x-2$ is a factor
of $ h(x)$, find the value of If $ k$ and hence solve the equation $ h(x)=0$.

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$\begin{aligned} & f(x)=4 x^{3}-4 k x^{2}-x+k \\\\ & g(x)=3 x^{2}+(1-3 k) x-k \\\\ & h(x)=f(x)+g(x) \\\\ & \therefore\ h(x)=4 x^{3}+(3-4 k) x^{2}-3 k x \\\\ & x-2 \text { is a factor of } h(x) . \\\\ &\therefore\ h(2)=0 \\\\ & 4(2)^{3}+(3-4 k)(2)^{2}-3 k(2)=0 \\\\ & 32+12-16 k-6 k=0 \\\\ & 44-22 k=0 \\\\ & \therefore\ k=2 \end{aligned}$