Question

Using factor theorem, determine whether g(x) is factor of p(x) in following cases: a) p(x) = x3 + 3x2 + 5x + 6, g(x) = x + 2
b) p(x) = 2x3 + x2
- 2x - 1, g(x) = x + 1

Answer 1

Answer :

Given :

♦ Cubic polynomial expressions & factors ;

➊

$\mathtt{\longrightarrow{p(x)={x}^{3} + 3{x}^{2} + 5x + 6}}$

$\mathsf{Factor: \: \: \: }{\tt{g(x) = x + 2}}$

➋

$\mathtt{\longrightarrow{p(x)=2{x}^{3} + {x}^{2} + 2x - 1}}$

$\mathsf{Factor: \: \: \: }{\tt{g(x) = x + 1}}$

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Required to find :

1. Whether the given factor is the factor of the expression or not ?

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Mentioned Condition :

• Solve using factor theorem

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Explanation :

Before solving this question we need to know some concept behind this question .

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What is a Factor Theorem ?

In algebra , Factor Theorem is a Theorem which is linking factors and zeros of the polynomial expression.

Actually, this factor theorem is a special case in the Remainder theorem .

This Theorem states that ,

For any given polynomial p(x) if (x - k) is a factor then p(k) gives the remainder as zero .

This means that k is the root of the expression .

For example :

$\green{\tt{If \; p(x) \; = 2{x}^{2}+x - 1\;\;\;\;and \; ( x + 1)\; is\; a\; factor}}$

$\blue{\textsf{So, equal the value of factor with zero in order to find the value of x }}$

$x + 1 = 0 \\ x = - 1$

So, value of x is - 1 .

Then substitute this value in p(x) instead of x .

$\orange{\tt{p(-1) = 2{(-1)}^{2} + (-1) - 1 }}$

$\orange{\tt{\Rightarrow{ 2(1) - 1 - 1 }}}$

$\orange{\tt{\Rightarrow{ 2 - 2}}}$

$\orange{\tt{\implies{ 0}}}$

Hence, (x - 1) is a factor of p(x) .

Using this let's solve this question .

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Solution :

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$\tt{p(x)= {x}^{3} + 3{x}^{2} + 5x +6\;\;,\;g(x)=x+2}$

Since, g(x) is a factor let's equal the value of g(x) with zero .

$\tt{x + 2 = 0}$

$\tt{x = - 2}$

Here, substitute this value in place of x in p(x)

$\tt{p(-2)= {(-2)}^{3} + 3{(-2)}^{2} + 5(-2) +6}$

$\tt{\Rightarrow{-8 + 3(4) - 10 + 6}}$

$\tt{\Rightarrow{- 8 + 12 - 10 + 6 }}$

$\tt{\Rightarrow{ -18 + 18}}$

$\tt{\red{\implies{0}}}$

Hence, g(x) is the factor of p(x) .

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$\tt{p(x)= 2{x}^{3} + {x}^{2} - 2x -1\;\;,\;g(x)=x+1}$

Since, g(x) is a factor let's equal the value of g(x) with zero .

$\tt{x + 1 = 0}$

$\tt{x = - 1}$

Here, substitute this value in place of x in p(x)

$\tt{p(-2)= 2{(-1)}^{3} + {(-1)}^{2} - 2(-1) - 1}$

$\tt{\Rightarrow{ 2(-1) + 1 + 2 - 1}}$

$\tt{\Rightarrow{ -2 + 1 + 2 - 1}}$

$\tt{\Rightarrow{ -3 +3}}$

$\tt{\red{\implies{0}}}$

Hence, g(x) is the factor of p(x) .

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