Ex.2.4 Q2 Polynomials Solution - NCERT Maths Class 9

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Question

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

 Video Solution
Polynomials
Ex 2.4 | Question 2

Text 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 \(p(x) = {x^3} - 4{x^2} + x + 6.\)

 Video Solution
Polynomials
Ex 2.4 | Question 2
  
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