Exercise 2.4 Polynomials NCERT Solutions Class 9

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Question 1

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

(i) \(\begin{align}{x}^{3}+{x}^{2}+x+1\end{align}\)

(ii) \(\begin{align}{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1\end{align}\)

(iii) \(\begin{align}{{x}^{4}}+3{{x}^{3}}+3{{x}^{2}}+x+1\end{align}\)

(iv) \(\begin{align}{{x}^{3}}-{{x}^{2}}-(2+\sqrt{2})x+\sqrt{2}\end{align}\)

Solution

Video Solution

Reasoning:

When a polynomial \(p(x)\) is divided by \(x-a\) and if \(p(a) = 0\) then \((x-a)\) is a factor of \(p(x)\). The root of \(x+1=0 \) is \( -1.\)

Steps:

(i) Let \(\begin{align} p(x)={{x}^{3}}+{{x}^{2}}+x+1\end{align} \)

\[\begin{align} \therefore p(-1)&={{(-1)}^{3}}+{{(-1)}^{2}}+(-1)+1 \\ & =-1+1-1+1=0 \\ \end{align} \]

Since the remainder of \( p\text{(-1) = 0} \) , we conclude that \((x+1)\) is a factor of \( {{x}^{3}}+{{x}^{2}}+x+1 \) .

(ii) Let \( p(x)={{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1 \)

\[\begin{align}\therefore p( - 1) &\!=\! {( - 1)^4}\!+\!{( - 1)^3}\!+\!{(\!-\!1)^2}\!+\!(\!-\!1)\!+\!1\\ &= \not 1 - \not 1 + \not 1 - \not 1+ 1\\ &= 1 \ne 0\end{align}\]

Since the remainder of  \( p( - 1) \ne 0\), we conclude that \((x+1)\) in not a factor of  \( \,{x^4} + {x^3} + {x^2} + x + 1\).

(iii) Let \( p(x) = {x^4} + 3{x^3} + 3{x^2} + x + 1\)

\[\begin{align} \therefore p( - 1) &\!=\!{( -\!1)^4}\!+\!3{(\!-\!1)^3}\!\!+\!3{(\!-\!1)^2}\!\!+\!( -\!1)\!\!+\!\!1 \\ &= 1 - 3 + 3 - 1 + 1 \\ &= 1 \ne 0\end{align}\]

Since the remainder of \(p( - 1) \ne 0\) ,  \((x+1)\) is not a factor of \({x^4} + 3{x^3} + 3{x^2} + x + 1\).

(iv) Let \(p(x) = {x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 \)

\[\begin{align} \therefore p( - 1) &\!=\!\!{(\!\!-\!1)^3}\!\!-\!{(\!-\!\!1)^2}\!\!-\!(2\!+\!\!\sqrt 2 )( -\!1)\!\!+\!\!\sqrt 2\!\\ &\!=\!-\!1\!-\!1\!+\!2 \!+ \!\sqrt 2\!+\!\sqrt 2\\ &= 2\sqrt 2 \end{align}\]

Since the remainder of \(p( - 1) = 0\) , \((x+1)\) is  not a factor of \({x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 \) .

Question 2

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

(i) \(p(x)\!=\!2{x^3}\!+\!{x^2}\!-\!2x\!-\!1,\,g(x)\!=\!x \!+\!1\)

(ii) \(p(x)\!=\!{x^3}\!+\!3{x^2}\!+\!3x\!+\!1,\,g(x)\!=\!x\!+\!2\)

(iii) \(p(x)\!=\!{x^3}\!-\!4{x^2}\!+\!x\!+\!6,\,g(x)\!=\!x\!-\!3\)

Solution

Video Solution

Reasoning:

By factor theorem, \((x-a)\) is a factor of a polynomial \( p(x)\) if \(p(a) = 0.\)

To find if \(g(x )= x+a\) is a factor of \( p(x),\) we need to find the root of \(g(x).\)

\(x + a = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = -a\)

Steps:

(i) Let \(p(x) = 2{x^3} + {x^2} - 2x - 1,\,g(x) = x + 1\)

\[x + 1 = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = {\rm{ }}-1\]

Now,

\[\begin{align}p( - 1) &= 2{( - 1)^3} + {( - 1)^2} - 2( - 1) - 1\\ & = - \not 2 + \not 1 + \not 2 - \not 1\\& = 0\end{align}\]

Since the remainder of \(p( - 1) = 0\) , by factor theorem we can say \(g(x) = x+1\) is a factor of \(p(x) = 2{x^3} + {x^2} - 2x - 1.\)

(ii) Let \(p(x) = {x^3} + 3{x^2} + 3x + 1,\,\,g(x) = x + 2\)

\[x + 2 = 0{\rm{ }} \to {\rm{ }}x{\rm{ }} = {\rm{ }}-2\]

Now,

\[\begin{align}p( - 2)&= {( - 2)^3} + 3{( - 2)^2} + 3( - 2) + 1\\ & = - 8 + 12 - 6 + 1\\ & = - 1 \ne 0\end{align}\]

Since the remainder of \(p( - 2) \ne 0\) , by factor theorem we can say \(g(x) = x+2\) is not a factor of \(p(x) = {x^3} + 3{x^2} + 3x + 1.\)

(iii) Let \(p(x) = {x^3} - 4{x^2} + x + 6,\,\,g(x) = x - 3\)

\[x - 3 = 0{\rm{ }} \to {\rm{ }}x\,{\rm{ = }}\,{\rm{3}}\]

Now,

\[\begin{align}p(3) &= {(3)^3} - 4{(3)^2} + 3 + 6\\ & = 27 - 36 + 3 + 6\\ & = 0\end{align}\]

Since the remainder of \(p(3) = 0\) , by factor theorem we can say \(g(x) = x-3\) is a factor of \(p(x) = {x^3} - 4{x^2} + x + 6.\)

Question 3

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

 

Solution

Video Solution

(i) \(\begin{align} p(x) = {x^2} + x + k\end{align}\)

(ii) \(\begin{align} p(x) = 2{x^2} + kx + \sqrt 2 \end{align}\)

(iii)  \(\begin{align} p(x) = k{x^2} - \sqrt {2x} + 1  \end{align}\) 

(iv)  \(\begin{align} p(x) = k{x^2} - 3x + k\end{align}\)

Reasoning:

By factor theorem, if \(x-1\) is a factor of \(p(x),\) then \( p\left( 1 \right) = 0\).

Steps:

(i) \(\begin{align} p(x) = {x^2} + x + k\end{align}\)

\[\begin{align}\,\,\,\, p(x) &= {x^2} + x + k\\ p(1) &= {(1)^2} + (1) + k\\ \,\,\,\,\,\,\,0\,\, &= 2 + k\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= - 2\end{align}\]

(ii) \(\begin{align} p(x) = 2{x^2} + kx + \sqrt 2 \end{align}\)

\[\begin{align}\,\,\,\, p(x) &= 2{x^2} + kx + \sqrt 2 \\ p(1) &= 2{(1)^2} + k(1) + \sqrt 2 \\ \,\,\,\,\,\,\,0\,\, &= 2 + k + \sqrt 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= - (2 + \sqrt 2 )\end{align}\]

(iii)  \(\begin{align} p(x) = k{x^2} - \sqrt {2x} + 1  \end{align}\)

\[\begin{align}\,\,\,\,\,\,\,\, p(x) &= k{x^2} - \sqrt {2x} + 1\\ p(1) &= k{(1)^2} - \sqrt {2(1)} + 1\\ \,\,\,\,\,\,\,\,0 &= k - \sqrt 2 + 1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= \sqrt 2 - 1\end{align}\]

(iv)  \(\begin{align} p(x) = k{x^2} - 3x + k\end{align}\)

\[\begin{align}\,\,\, p(x) &= k{x^2} - 3x - k\\ p(1) &= k({1^2}) - 3(1) - k\\ \,\,\,\,\,\,\,\,0 &= 2k - 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= \frac{3}{2}\end{align}\]

Question 4

Factorise:

(i)\(\begin{align}12{x^2} - 7x + 1 \end{align}\)

(ii)\(\begin{align}2{x^2} + 7x + 3 \end{align}\)

(iii) \(\begin{align}6{x^2} + 5x - 6 \end{align}\)

(iv) \(\begin{align}3{x^2} - x - 4 \end{align}\)

Solution

Video Solution

Reasoning:

By splitting method, we can find factors using the following method.

Find \(2\) numbers \(p, q\) such that:

(i) \(p + q = \) co-efficient of \(x\)

(ii) \(pq = \) co-efficient of \({x^2}\) and the constant term.

Steps:

(i) \(\,12{x^2} - 7x + 1\)

\(p + q = - 7\)(co-efficient of \(x\))

\(pq = 12 \times 1 = 12\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = -4, q = -3.\)

Now splitting the middle term of the given polynomial,

\[\begin{align}12{x^2} - 7x + 1 &= 12{x^2} - 4x - 3x + 1\\ &= 4x(3x - 1) - 1(3x - 1)\\ &= (3x - 1)(4x - 1) \\&\text {(taking} (3x\!-\!1\!)\!\text{ as common)}\end{align}\]

(ii) \(\,\,2{x^2} + 7x + 3\)

\(p + q = 7\) (co-efficient of \(x\))

\(pq = 2 \times 3 = 6\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = 6, q = 1.\)

Now splitting the middle term of the given polynomial,

\[\begin{align}\,2{x^2} + 7x + 3 &= 2{x^2} + 6x + x + 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 2x(x + 3) + 1(x + 3)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (2x + 1)(x + 3)\end{align}\]

(iii) \(6{x^2} + 5x - 6\)

\(p + q = 5\) (co-efficient of \(x\))

\(pq = 6 \times -6 = 36\) (co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = 9, q = -4.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 6{x^2} + 5x - 6 &= 6{x^2} + 9x - 4x - 6\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 3x(2x + 3) - 2(2x + 3)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (3x - 2)(2x + 3)\end{align}\]

(iv) \(3{x^2} - x - 4\)

\(p + q = -1\) (co-efficient of \(x\))

\(pq = 3 \times -4 = -12\)(co-efficient of \({x^2}\,\) and the constant term.)

By trial and error method, we get \(p = -4, q = 3.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 3{x^2} - x - 4 &= 3{x^2} - 4x + 3x - 4\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \,\,3{x^2} + 3x - 4x - 4\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \,3x(x + 1) - 4(x + 1)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (3x - 4)(x + 1)\end{align}\]

Question 5

Factorise:

(i) \(\begin{align}{x^3} - 2{x^2} - x + 2 \end{align}\)

(ii) \(\begin{align}{x^3} - 3{x^2} - 9x - 5\end{align}\)

(iii) \(\begin{align}{x^3} + 13{x^2} + 32x + 20 \end{align}\)

(iv) \(\begin{align}2{y^3} + {y^2} - 2y - 1\end{align}\)

Solution

Steps:

(i) Let \(p(x) = {x^3} - 2{x^2} - x + 2\)

By the factor theorem we know that \(x-a\) is a factor of \(p(x) \)if \(p(a) = 0.\)

We shall find a factor of \(p(x) \) by using some trial value of \(x,\) say \(x = 1.\)

\[\begin{align} p(1) &= {(1)^3} - 2{(1)^2} - 1 + 2\\ &= 1 - 2 - 1 + 2 = 0\end{align}\]

Since the remainder of \(p(1) = 0\) , by factor theorem we can say \(x=1\) is a factor of \(p(x) = {x^3} - 2{x^2} - x + 2.\)

Now divide \(p(x)\) by \(x-1\) using long division,

Hence \({x^3}\!-\!\! 2{x^2}\! -\! x\!+\!\!2\!=\!(x\!-\!\!1)({x^2}\!-\!x\!-\!\!2)\)

Now taking \({x^2} - x - 2\) , find \(2\) numbers \(p, q\) such that:

(i) \(p + q = \)co-efficient of \(x\)

(ii) \( pq = \) co-efficient of \({x^2}\) and the constant term.

\(p + q = - 1\) (co-efficient of \(x\))

\(pq = 1 \times - 2 = - 2\)(co-efficient of \({x^2}\) and the constant term.)

By trial and error method, we get \(p = -2, q = 1.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} {x^2} - x - 2 &= {x^2} - 2x + x - 2\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= x(x - 2) + 1(x - 2)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (x + 1)(x - 2)\end{align}\]

\( \therefore\!{x^3}\!\!- \!2{x^2}\!\!-\!x\!+\!2\!=\!(x\!\!-\!1)\!(x\!-\!2)(x\!+\!\!1)\)

Method 2:

\[\begin{align}{x^3}\!\!-\!2{x^2}\!\!-\!x\!+\!2\!&=\!({x^3}\!\!-\!2{x^2})\!-\!(x\!\!-\!\!2)\\ &= {x^2}(x - 2)\!-\!1(x\! -\!\! 2)\\ &= (x - 2)({x^2} - 1)\\ &= (x\!-\!2)(x\!+\!1)(x\!-\!\! 1) \\&\left( \begin{array}{l}{\text{By using }}{a^2} - {b^2}\\ = (a + b)(a - b)\end{array} \right)\end{align}\]

(ii) Let \(p(x) = {x^3} - 3{x^2} - 9x - 5\)

By the factor theorem we know that \(x-a\) is a factor of \(p(x)\) if \(p(a) = 0.\)

We shall find a factor of \(p(x)\) by using some trial value of \(x,\) say \(x = 1.\)

\[\begin{align}p(1) &= {(1)^3} - 3{(1)^2} - 9(1) - 5 \\ &= 1 - 3 - 9 - 5\\ &= - 16 \ne 0\end{align}\]

Since the remainder of \(p(1) \ne 0\) , by factor theorem we can say \(x=1\) is not a factor of \(p(x) = {x^3} - 3{x^2} - 9x - 5.\)

Now say \(x = -1.\)

\[\begin{align} p(\!-\!1) &\!=\!\!{(\!-\!1)^3}\!-\!\!3{( -\!1)^2}\!\!-\!\! 9(\!-\!1)\!-\!\!5 \\ &= - 1 - 3 + 9 - 5\\ &= - 9 + 9 = 0\end{align}\]

Since the remainder of \(p( - 1) = 0\) , by factor theorem we can say \(x=-1\) is a factor of \(p(x) = {x^3} - 3{x^2} - 9x - 5.\)

Now dividing \(p(x)\) by \( x+1\)using long division.

Hence \({x^3} -\!\!3{x^2}\!\!-\!\!9x\!-\!\!5\!\!=\!\!(x\!+\!1)\!({x^2}\!-\!4x\!-\!5)\)

Now taking \({x^2} - 4x - 5\) , find \(2\) numbers \(p, q\) such that:

(i) \(p + q =\) co-efficient of \(x\)

(ii) \(pq =\) co-efficient of \({x^2}\) and the constant term.

\(p + q = - 4\) (co-efficient of \(x\))

\(pq = 1 \times - 5 = - 5\) (co-efficient of \({x^2}\) and the constant term.)

By trial and error method, we get \(p = -5, q = 1.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} {x^2} - 4x - 5 &= {x^2} - 5x + x - 5\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= x(x - 5) + 1(x - 5)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= (x + 1)(x - 5) \end{align}\]

\[\begin{align}&\therefore {x^3}\!-\!\!2{x^2}\!-\!\!x\!+\!2 \\&= (x\!+\!1)(x\!-\!5)(x\!+\!1)\\ &= \,{(x\!+\!1)^2}\!(x\!- \!\!5)\,\, \end{align}\]

(iii) Let \(p(x) = {x^3} + 13{x^2} + 32x + 20\)

By the factor theorem we know that \( x-a\) is a factor of \(p(x)\) if \(p(a) = 0.\)

We shall find a factor of \(p(x)\) by using some trial value of \(x,\) say \(x = -1.\) (Since all the terms are positive.)

\[\begin{align}\!p(\!-\!1) &\!=\!{(\!- 1)^3}\!+\!\!13{( - 1)^2}\!\!+\!32( - 1)\!+\!\!20 \\ &= - 1 + 13 - 32 + 20\\ &= 0 \end{align}\]

Since the remainder of \(p( - 1) = 0\) , by factor theorem we can say \(x= -1\) is a factor of \(p(x) = {x^3} + 13{x^2} + 32x + 20.\)

Now dividing \(p(x)\) by \(x+1\) using long division,

\[\left[ \begin{array}{l}\therefore {x^3} + 13{x^2} + 32x + 20\\ = \,(x + 1)({x^2} + 12x + 20)\end{array} \right]\]

Now taking \({x^2} + 12x + 20\) , find \(2\) numbers \(p, q\) such that:

(i) \(p + q =\) co-efficient of \(x\)

(ii) \(pq =\) co-efficient of \({x^2}\) and the constant term.

\(p + q = 12\) (co-efficient of \(x\))

\(pq = 1 \times 20 = 20\) (co-efficient of \({x^2}\) and the constant term.)

By trial and error method, we get \(p = 10, q = 2.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} {x^2}\!\!+\!12x\!+\!20\!&=\!{x^2}\!\!+\!10x\!+\!\!2x\!+\!\!20\\ &=\!x(x\!+\!\!10)\!+\!\!2(x\!+\!\!10)\\ &= (x\!+\!\!10)\!(x\!+\!\!2)\end{align}\]

\[\left( \begin{array}{l}\therefore {x^3} + \,13{x^2} + \,32x\, + \,20\\ = (x + 1)\,(x + 10)\,(x + 2)\end{array} \right)\]

 Method 2:

\[\begin{align}&{x^3} + 13{x^2} + 32x + 20 \\&=\!{x^3}\!+\!10{x^2}\!+\!3{x^2}\!+\!30x\!+\!2x\!+\!20\\ &= {x^2}\!(x\!+\!\!10)\!+\!\!3x(x\!+\!10)\!+\!\!2(\!x\!+\!10\!) \\ &= (x + 10)({x^2} + 3x + 2)\\ &= (x + 10)({x^2} + 2x + x + 2) \\ &= (x + 10)[x(x + 2) + 1(x + 2)]\\ &= (x + 10)(x + 2)(x + 1)\end{align}\]

(iv) Let  \(p(y) = 2{y^3} + {y^2} - 2y - 1\)

By the factor theorem we know that \((y-a)\) is a factor of \(p(y)\) if \(p(a) = 0.\)

We shall find a factor of \(p(y)\) by using some trial value of \(y,\) say \(y = 1.\)

\(\begin{align} p(1) &= 2{(1)^3} + {(1)^2} - 2(1) - 1\\ \,\,\,\,\,\,\,\,\,\,\,\, &= 2 + 1 - 2 - 1\\ \,\,\,\,\,\,\,\,\,\,\,\, &= 0 \end{align}\)

Since the remainder of \(p(1) = 0\) , by factor theorem we can say \(y-1\) is a factor of \(p(y) = 2{y^3} + {y^2} - 2y - 1\)

Now dividing \(p(y)\) by \(y-1\) using long division,

\(\therefore\!2{y^3}\!+\!{y^2}\!-\!2y\!-\!\!1\!\!=\!\!(y\!\!-\!\!1)\!(2{y^2}\!+\!3y\!+\!1)\)

Now taking \(2{y^2} + 3y + 1\) , find \(2\) numbers \(p, q\) such that:

(i) \(p + q = \)co-efficient of \(y\)

(ii) \(pq =\) co-efficient of \({y^2}\) and the constant term.

\(p + q = 3\) (co-efficient of \(y\))

\(pq = 2 \times 1 = 2\)(co-efficient of \({y^2}\) and the constant term.)

By trial and error method, we get \(p = 2, q = 1.\)

Now splitting the middle term of the given polynomial,

\[\begin{align} 2{y^2} + 3y + 1 &= 2{y^2} + 2y + y + 1 \\ &= 2y(y + 1) + 1(y + 1)\\&= (2y + 1)(y + 1)\end{align}\]

\(\!\therefore\!2{y^3}\!+\!\!{y^2}\!-\!2y\!-\!\!1\!\!=\!\!(y\!-\!1)(2y\!+\!\!1)\!(y\!+\!\!1)\)

  
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