Polynomials Exercise 2.3 Solutions – NCERT Class 9 Maths Chapter 2

1. Determine which of the following polynomials has (x + 1) a factor :

Solution:
zero of $\left( x+1\right)=-1$

$p\left( x\right) =x^{3}+x^{2}+x+1\\$
$$\begin{aligned} p\left( -1\right) &&=\left( -1\right) ^{3}+\left( -1\right) ^{2}-1+\\&&=-1+1-1+1\\&&=0\end{aligned} $$ $\therefore x+1$ is a factor of $p\left( x\right) =x^{3}+x^{2}+x+1$

$p\left( x\right) =x^{4}+x^{3}+x^{2}+x+1$
$$\begin{aligned} p\left( -1\right) &=\left( -1\right) ^{4}+\left( -1\right) ^{3}+\left( -1\right) ^{2}-1+1\\ &=1-1+1-1+1\\ &=3-2\\ &=1\end{aligned}$$ $\therefore x+1$ is not a factor of $p\left( x\right) =x^{4}+x^{3}+x^{2}+x+1$

$p\left( x\right) =x^{4}+3x^{3}+3x^{2}+x+1$
$$\begin{aligned} p\left( -1\right) &=\left( -1\right) ^{4}+3\left( -1\right) ^{3}+3\left( -1\right) ^{2}-1+1\\ &=1-3\left( -1\right) +3\left( 1\right) -1+1\\ &=1+3+3-1+1\\ &=8-1\\ &=7\end{aligned}$$ $\therefore x+1$ is not a factor of $p\left( x\right) =x^{4}+3x^{3}+3x^{2}+x+1$

$p\left( x\right)=x^{3}-x^{2}-\left( 2+\sqrt{2}\right) x+\sqrt{2}$ $$\begin{aligned} p\left( -1\right) &=\left( -1\right) ^{3}-\left( -1\right) ^{2}-\left( 2+\sqrt{2}\right) \left( -1\right) +\sqrt{2}\\ &=-1-1+\left( 2+\sqrt{2}\right) +\sqrt{2}\\ &=-2+2+\sqrt{2}+\sqrt{2}\\ &=2\sqrt{2}\end{aligned}$$ $\therefore x+1$ is not a factor of $p\left( x\right)=x^{3}-x^{2}-\left( 2+\sqrt{2}\right) x+\sqrt{2}$

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

$$\begin{aligned} g\left( x\right) &=x+1\\ g\left( 0\right) \Rightarrow x&=-1\\ P\left( -1\right) &=2\left( -1\right) ^{3}+\left( -1\right) ^{2}-2\left( -1\right) -1\\ &=-2+1+2-1\\ &=3-3\\ &=0\end{aligned}$$ $\therefore g(x)$ is a factor of $p(x)$

$$\begin{aligned}p\left( x\right) &=x^{3}+3x^{2}+3x+1\\ g\left( x\right) &=x+2\\ g\left( 0\right) \Rightarrow x&=-2\\ p\left( -2\right) &=\left( -2\right) ^{3}+3\left( -2\right) ^{2}+3\left( -2\right) +1\\ &=-8+3\times 4-6+1\\ &=-8+12-6+1\\ &=13-14\\ &=-1\end{aligned}$$ $\therefore g(x)$ is not a factor of $p(x)$

$$\begin{aligned}p\left( x\right) &=x^{3}-4x^{2}+x+6\\ g\left( x\right) &=x-3\\ g\left( 0\right) \Rightarrow x&=3\\ p\left( 3\right) &=\left( 3\right) ^{3}-4\left( 3\right) ^{2}+3+6\\ &=27-4\times 9+3+6\\ &=27-36+9\\ &=36-36\\ &=0\end{aligned}$$ $\therefore g(x)$ is a factor of $p(x)$

3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

$p\left( x\right) =x^{2}+x+k$ $$\begin{aligned} p\left( 1\right) &=\left( 1\right) ^{2}+1+k\\ 0&=1+1+k\\ \Rightarrow k+2&=0\\ k&=-2\end{aligned}$$

$\left( x\right) =2x^{2}+kx+\sqrt{2}\\$ $$\begin{aligned}p P\left( 1\right) &=2\cdot 1^{2}+k\cdot 1+\sqrt{2}\\ 0&=2+k+\sqrt{2}\\ 0&=k+\left( 2+\sqrt{2}\right) \\ \Rightarrow k+\left( 2+\sqrt{2}\right) &=0\\ k&=-\left( 2+\sqrt{2}\right) \end{aligned}$$

$p\left( x\right) =kx^{2}-\sqrt{2}x+1\\$ $$\begin{aligned} p\left( 1\right) &=k\left( 1\right) ^{2}-\sqrt{2}\cdot 1+1\\ 0&=k-\sqrt{2}+1\\ \Rightarrow k-\sqrt{2}+1&=0\\ \Rightarrow k&=\sqrt{2}-1\end{aligned}$$
$p\left( x\right) =kx^{2}-3x+k\\$ $$\begin{aligned} p\left( 1\right) &=k\left( 1\right) ^{2}-3\left( 1\right) +k\\ 0&=k-3+k\\ 0&=2k-3\\ \Rightarrow 2k-3&=0\\ 2k&=3\\ k&=3/2\end{aligned}$$

4. Factorise :

$$\begin{aligned}12x^{2}-7x+1\\ &=12\left( x^{2}-\dfrac{7}{12}x+\dfrac{1}{12}\right) \\ &=12p\left( x\right) \\ p\left( x\right) &=x^{2}-\dfrac{7}{12}+\dfrac{1}{12}\quad\color{blue}\Rightarrow\left( x+a\right) \left( x+b\right) =x^{2}+\left( a+b\right) x+ab\end{aligned}$$ $\text{Comparing constant term with \(p(x)\Rightarrow$} ab=\frac{1}{12}\) $$ Let~us~find~the~possible~values~of a~ and ~b~which~gives~us ~ab=\frac{1}{12} ~ and~ a+b=\frac{7}{12}$$ $$\begin{aligned} \left( a,b\right) &=\left( \pm 1,\pm \frac{1}{12}\right) ,\left( \pm \dfrac{1}{2},\pm \dfrac{1}{6}\right) ,\left( \pm \dfrac{1}{3},\pm \dfrac{1}{4}\right)\end{aligned}$$ $$\text{Among these values only } \left( \pm \dfrac{1}{3},\pm \dfrac{1}{4}\right)\text{ can give us required result when $p(x)$ may be zero }$$ $$\begin{aligned} p\left( x\right) &=x^{2}-\dfrac{7}{12}x+\dfrac{1}{12}\\ p\left( \frac{1}{3}\right) &=\left( \frac{1}{3}\right) ^{2}-\dfrac{7}{12}\times \dfrac{1}{3}+\dfrac{1}{12}\\ &=\dfrac{1}{9}-\dfrac{7}{36}+\dfrac{1}{12}\\ &=\dfrac{4-7+3}{36}\\ &=\dfrac{7-7}{36}\\ &=0\\ \color{blue}\therefore\quad \left( x-1/3\right)\text{ is a factor of } p(x) \end{aligned}$$ $$\require{cancel}\begin{aligned}p\left( \frac{1}{4}\right) &=\left( \frac{1}{4}\right) ^{2}-\dfrac{7}{12}\left( \frac{1}{4}\right) +\dfrac{1}{12}\\ &=\dfrac{1}{16}-\dfrac{7}{48}+\dfrac{1}{12}\\ &=\dfrac{3-7+4}{48}\\ &=\dfrac{7-7}{48}\\ &=0\\ \color{blue}\therefore\quad \left( x-1/4\right) \text{is another factor of }p\left( x\right) \\ \therefore\quad p\left( x\right) &=\left( x-\frac{1}{3}\right) \left( x-\frac{1}{4}\right) \\ 12\cdot p\left( x\right) &=12\left( x-\frac{1}{3}\right) \left(x- \frac{1}{4}\right) \\ &=12\left( \dfrac{3x-1}{3}\right) \left( \dfrac{4x-1}{4}\right) \\ &=\dfrac{\cancel{12}}{\cancel{12}}\left( 3x-1\right) \left( 4x-1\right) \\ &=\left( 3x-1\right) \left( 4x-1\right) \end{aligned}$$

$2x^{2}+7x+3\\$ $$\begin{aligned}2x^{2}+7x+3\\ &=2\left( x^{2}+\dfrac{7}{2}x+\dfrac{3}{2}\right) \\ &=2p\left( x\right) \\ p\left( x\right) &=x^{2}+\dfrac{7}{2}x+\dfrac{3}{2}\quad\Rightarrow\color{blue}\left( x+a\right) \left( x+b\right) =x^{2}+\left( a+b\right) x+ab\\\end{aligned}$$ $$\text{ Let us find all possible values of } a ~ and ~ b \text{ such that } a.b=3/2 ~and~ a + b=7/2\left( a,b\right)$$ $$\begin{aligned} \left( a,b\right) &=\left( \pm 1,\pm \frac{3}{2}\right) ,\left( \pm \frac{1}{2},\pm 3\right) \\ &=\frac{1}{2},\pm 1,\pm \frac{3}{2},\pm 3\\ p\left( x\right) &=x^{2}+\dfrac{7x}{2}+\dfrac{3}{2}\\\\ p\left( \frac{-1}{2}\right) &=\left( -\frac{1}{2}\right) ^{2}+\dfrac{7x}{2}\left( \frac{-1}{2}\right) +\frac{3}{2}\\ &=\dfrac{1}{4}-\dfrac{7}{4}+\dfrac{3}{2}\\ &=\dfrac{1-7+6}{4}\\ &=\dfrac{7-7}{4}\\ &=0\\ \color{blue}\therefore\quad \left( x+\dfrac{1}{2}\right) is~a~factor~of~p\left( x\right) \\ p\left( x\right) &=x^{2}+\dfrac{7x}{2}+\dfrac{3}{2}\\ P\left( -1\right) &=\left( -1\right) ^{2}+\dfrac{7x}{2}\left( -1\right) +\dfrac{3}{2}\\ &=1-\dfrac{7}{2}+\dfrac{3}{2}\\ &=\dfrac{2-7+3}{2}\\ &=\dfrac{5-7}{2}\\ &=-\dfrac{3}{2}\\ &\neq 0\\ p\left( \frac{-3}{2}\right) &=\left( \frac{-3}{2}\right) ^{2}+\dfrac{7}{2}\left( \frac{-3}{2}\right) +\dfrac{3}{2}\\ &=\dfrac{9}{4}-\dfrac{21}{4}+\dfrac{3}{2}\\ &=\dfrac{\left( 9-21+6\right) }{4}\\ &=\dfrac{15-21}{4}\\ &=-\dfrac{6}{4}\\ &\neq 0\end{aligned}$$ $$\begin{aligned}p\left( -3\right) &=\left( -3\right) ^{2}+\dfrac{7}{2}\left( -3\right) +\dfrac{3}{2}\\ &=9-\dfrac{7}{2}\times 3+\dfrac{3}{2}\\ &=9-\dfrac{21}{2}+\dfrac{3}{2}\\ &=\dfrac{18-21+3}{2}\\ &=\dfrac{21-21}{2}\\ &=0\\ \Rightarrow \left( x+3\right) ~is~another~factor\\ p\left( x\right) &=\left( x+\dfrac{1}{2}\right) \left( x+3\right) \\ 2p\left( x\right) &=\dfrac{2\left( 2x+1\right) \left( x+3\right) }{2}\\ 2x^{2}+7x+3&=\left( 2x+1\right) \left( x+3\right) \end{aligned}$$ By mid-term Splitting Method: $$2x^{2}+7x+3$$ $$p+q=7 ~\&~ pq =6 $$ If we can find two numbers p and q such that then we can factorise the polynomial pq = 6, possible value of p and q (P, q) $\Rightarrow$ (1,6), (2, 3) pair (1,6) can give $$p+q=7$$ $$\begin{aligned}2x^{2}+7x+3\\ &=2x^{2}+x+6x+3\\ &=x\left( 2x+1\right) +3\left( 2x+1\right) \\ &=\left( 2x+1\right) \left( x+3\right) \end{aligned}$$

$6x^{2}+5x-6\\$ Let us find two numbers $p$ and $q$ such that $p\cdot q=36$ and $p+q=5$ $$\left( p,q\right) =\left( 1,36\right) ,\left( 2,18\right) ,\left( 3,12\right) ,\left( 4,9\right) ,\left( 6,6\right) $$ $$\begin{aligned}~of~these~4,9~will~give~us~p-q&=5~so,\\ 6x^{2}+5x-6&=6x^{2}+\left( 9-4\right) x-6\\ &=6x^{2}+9x-4x-6\\ &=3x\left( 2x+3\right) -2\left( 2x+3\right) \\ &=\left( 2x+3\right) \left( 3x-2\right) \end{aligned}$$

$3x^{2}-x-4\\$ Let us find values-of p and q, such that $$\begin{aligned}p+q&=-1\\ pq&=-12\end{aligned}$$ Let us find possible values of p q $$\begin{aligned}\left( p,q\right) &=\left( \pm 1,\pm 12\right) ,\left( \pm 2,\pm 6\right) ,\left( \pm 3,\pm 4\right) \\ \left( p,q\right) &=\left( 3,-4\right) \\\\ 3x^{2}-x-4 &=3x^{2}+3x-4x-4\\ &=3x\left( x+1\right) -4\left( x+1\right) \\ &=\left( x+1\right) \left( 3x-4\right) \end{aligned}$$ By Factor Theorem: $$\begin{aligned}3x^{2}-x-4 &=3\left( x^{2}-\dfrac{x}{3}-\dfrac{4}{3}\right) \\ &=3\cdot p\left( x\right) \\ p\left( x\right) &=x^{2}-\dfrac{x}{3}-\dfrac{4}{3}\quad\Rightarrow\color{blue}\left( x+a\right) \left( x+b\right) =x^{2}+\left( a+b\right) x+ab\\ ab&=-4/3\end{aligned}$$ Let us find possible values of $a~and~b$ by comparing constant value of $p(x)$ $$\begin{aligned}\left( a,b\right) &=\left( \pm 1,\pm \frac{4}{3}\right) ,\left( \pm 2,\pm \dfrac{2}{3}\right) \\ &=\pm 1,\pm \frac{4}{3},\pm \frac{2}{3},\pm 2\\ p\left( -1\right) &=1^{2}-\dfrac{1}{3}\left( -1\right) -\dfrac{4}{3}\\ &=1+\dfrac{1}{3}-\dfrac{4}{3}\\ &=\dfrac{3+1-4}{3}\\ &=\dfrac{4-4}{3}\\ &=0\\ \color{blue}\therefore\quad \left( x+1\right) \text{ is a factor of }p(x)\end{aligned}$$ $$\begin{aligned}p\left( \frac{-4}{3}\right) &=\left( \dfrac{4}{3}\right) ^{2}-\dfrac{1}{3}\left( -\dfrac{4}{3}\right) -\dfrac{4}{3}\\ &=\dfrac{16}{9}+\dfrac{4}{9}-\dfrac{4}{3}\\ &=\dfrac{16+4-12}{9}\\ &=\dfrac{-8}{9}\neq 0\\ P\left( \dfrac{4}{3}\right) &=\left( \dfrac{4}{3}\right) ^{2}-\dfrac{1}{3}\left( \dfrac{4}{3}\right) -\dfrac{4}{3}\\ &=\dfrac{16}{9}-\dfrac{4}{9}-\dfrac{4}{3}\\ &=\dfrac{16-4-12}{9}\\ &=\dfrac{16-16}{9}\\ &=0\\ \therefore\quad\color{blue} \left( x-\frac{4}{3}\right)\text{ is another factor of }p(x) \end{aligned}$$ $$\require{cancel}\begin{aligned}p\left( x\right) &=\left( x+1\right) \left( x-\dfrac{4}{3}\right) \\ &=\dfrac{\left( x+1\right) \left( 3x-4\right) }{3}\\ 3\cdot p\left( x\right) &=\cancel{3}\cdot\frac{\left( x+1\right) \left( 3x-4\right)}{\cancel{3}} \\ &=\left( x+1\right) \left( 3x-4\right) \end{aligned}$$

5. Factorise :

$x^{3}-2x^{2}-x+2\\$ $$\begin{aligned}x^{3}-2x^{2}-x+2 &=x^{2}\left( x-2\right) -1\left( x-2\right) \\ &=\left( x-2\right) \left( x^{2}-1\right) \\ &=\left( x-2\right) \left( x-1\right) \left( x+1\right) \end{aligned}$$

$x^{3}-3x^{2}-9x-5\\$ $$ \begin{aligned} x^{3}-3x^{2}-9x-5\\ \text{Let's find possible factors of 5} \pm 1,\pm\\ Let~p\left( x\right) &=x^{3}-3x^{2}-9x-5\\ p\left( -1\right) &=\left( -1\right) ^{3}-3\left( -1\right) ^{2}-9\left( -1\right) -5\\ &=-1-3+9-5\\&=9-9\\&=0\\ \therefore\quad\color{blue}\left( x+1\right)\text{ is a factor of }p(x)\\ \\ p\left( 1\right) &=1^{3}-3\cdot 1^{2}-9\cdot 1-5\\ &=1-4-9-5\\ &=-17\\ &\neq 0\\ p\left( -5\right) &=\left( -5\right) ^{3}-3\left( -5\right) ^{2}-9\left( -5\right) -5\\ &=125-3\times \left( -25\right) +45-5\\ &=125+75+45-5\\ &=240\\ &\neq 0\\ p\left( 5\right) &=5^{3}-3\cdot 5^{2}-9\left( 5\right) -5\\ &=125-3\times 25-45-5\\ &=125-75-50\\ &=0\\ \therefore\quad\color{blue}\left( x-5\right) ~is~another~factor\\\\ (x+1)(x-5)&=x^2-4x-5\\ \frac{x^{3}-3x^{2}-9x-5}{x^2-4x-5}&=(x+1)\\\\ \require{enclose} x+1\phantom{0000000}\\ x^2 -4x-5\enclose{longdiv}{x^{3}-3x^{2}-9x-5}\\ \underline {\_x^3-\_4x^2-\_5x}\phantom{0000}\\x^2-4x-5\phantom{0}\\ \underline{{\_x^2-\_4x-\_5}}\phantom{0}\\ \text{x }\quad\text{ x }\quad\text{ x}\phantom{0}\\\\ \color{blue}\therefore\quad (x+1) \text{is yet another factor of } p(x)\\ \therefore\quad x^{3}-3x^{2}-9x-5&=\left( x-5\right) \left( x+1\right) \left( x+1\right)\\ \end{aligned} $$
$x^{3}+13x^{2}+32x+20\\$ $$\begin{aligned}x^{3}+13x^{2}+32x+20\\ \text{Let's find possible factors of 20 }\\ \left( \pm 1,\pm 20\right) ,\left( \pm 2,\pm 10\right) ,\left( \pm 4,\pm 5\right) \\ p\left( x\right) &=x^{3}+13x^{2}+32x+0\\ P\left( -1\right) &=\left( -1\right) ^{3}-13\left( -1\right) ^{2}-32\left( -1\right) +20\\ &=-1+13-32+20\\ &=33-33\\ &=0\\ \color{blue}\therefore\quad \left( x+1\right) \text{ is a factor of }p(x)\\p\left( -2\right) &=\left( -2\right) ^{3}+13\left( -2\right) ^{2}+32\left( -2\right) +20\\ &=-8+13\times 4-32\times 2+2\\&=-8+52-64+20\\ &=-72+72\\ &=0\\ \color{blue}\therefore\quad\left( x+2\right)\text{is another factor of } p(x)\\ p\left( -4\right) &=\left( -4\right) ^{3}+13\left( -4\right) ^{2}+32\left( -4\right) +20\\ &=-64+13\times 16-128+20\\ &=-64+208-128+20\\ &=-192+228\\ &=36\\ &\neq 0\\ p\left( -5\right) &=\left( -s\right) ^{3}+13\left( -5\right) ^{2}+32\left( -5\right) +20\\ &=-125+13\times 25-160+20\\ &=-125+325-160+20\\ &=60\\ &\neq 0\\p\left( -10\right) &=\left( -10\right) ^{3}+13\left( -10\right) ^{2}+32\left( -10\right) +20\\ &=-1000+1300-320+20\\ &=-1320+1320\\ &=0\\ \color{blue}\therefore\quad x+10\text{ is yet another factor of }p(x)\\\\ x^{3}+13x^{2}+32x+2&=\left( x+1\right) \left( x+2\right) \left( x+10\right) \end{aligned} $$
$2y^{3}+y^{2}-2y-1\\$ $$\require{cancel} \begin{aligned}2y^{3}+y^{2}-2y-1 &=2\left( y^{2}+\dfrac{y^{2}}{2}-y-1/2\right) \\ 2p\left( y\right) &=y^{3}+\dfrac{y^{2}}{2}-y-\frac{1}{2}\\ p\left( y\right) &=y^{3}+\dfrac{y^{2}}{2}-y-\frac{1}{2}\\ p\left( -1\right) &=\left( -1\right) ^{3}+\dfrac{\left( -1\right) ^{2}}{2}-\left( -1\right) \frac{-1}{2}\\ &=-1+\frac{1}{2}+1-\frac{1}{2}\\ &=0\\ \color{blue}\therefore\quad y+1 \text{ is a factor of }p(y)\\ p\left( 1\right) &=\left( 1\right) ^{3}+\dfrac{1^{2}}{2}-1-1,2\\ &=1+\frac{1}{2}-1\frac{-1}{2}\\ &=0\\ \color{blue}\therefore\quad \left( y-1\right)\text{ is another factor of }p(y) \\ p\left( \frac{1}{2}\right) &=\left( \frac{1}{2}\right) ^{3}+\dfrac{\left( 1/2\right) ^{2}}{2}-1/2\frac{-1}{2}\\ &=\dfrac{1}{8}+\dfrac{1}{4\cdot 2}-1/2\frac{-1}{2}\\ &=\dfrac{1}{8}+\dfrac{1}{8}-1\\ &=\dfrac{1}{4}-1\\ &=-3/4\\ &\neq 0\\\\ p\left( -\frac{1}{2}\right) &=\left( \frac{-1}{2}\right) ^{3}+\dfrac{\left( \frac{-1}{2}\right) ^{2}}{2}-\left( -1/2\right) -1/2\\ &=\dfrac{-1}{8}+\dfrac{1}{8}+\frac{1}{2}\frac{-1}{2}\\ &=0\\ \color{blue}\therefore\quad \left( y+\frac{1}{2}\right)\text{ is yet another factor of }p(y) \\ p\left( y\right) &=\left( y-1\right) \left( y+1\right) \left( \dfrac{2y+1}{2}\right) \\ &=\dfrac{\left( y-1\right) \left( y+1\right) \left( 2y+1\right) }{2}\\ 2p\left( y\right) \Rightarrow \cancel{2}\dfrac{\left( y-1\right) \left( y+1\right) \left( 2y+1\right) }{\cancel{2}}&=\left( y-1\right) \left( y+1\right) \left( 2y+1\right) \end{aligned}$$