Ex.2.4 Q5 Polynomials Solution - NCERT Maths Class 9


Question

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

Text 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,

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

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

 Video Solution
Polynomials
Ex 2.4 | Question 5
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