How do you use the factor theorem to determine whether $3x+1$ is a factor of $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-55{{x}^{2}}+163x+60$ ?
Answer
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Hint: Here we have to use the factor theorem to determine whether the given expression is a factor of the polynomial given. Firstly we will write the factor theorem and the necessary condition for an expression to be the factor of a polynomial. Then we will simplify accordingly and determine whether the condition is satisfied or not and get our desired answer.
Complete step by step answer:
We have to determine whether $3x+1$ is a factor of the below polynomial:
$f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-55{{x}^{2}}+163x+60$….$\left( 1 \right)$
Now as we know that factor theorem states that if we have a polynomial $f\left( x \right)$ of degree$n\ge 1$ where $a$ is any real number than $\left( x-a \right)$ is a factor of the polynomial if the condition $f\left( a \right)=0$ is satisfied.
So as we have been given the expression as follows:
$3x+1$
$\Rightarrow 3x+1=0$
So we get,
$\Rightarrow 3x=-1$
$\Rightarrow x=-\dfrac{1}{3}$
So we get $x=-\dfrac{1}{3}$ substitute it in equation (1) as follows:
$\Rightarrow f\left( \dfrac{-1}{3} \right)=3{{\left( \dfrac{-1}{3} \right)}^{4}}-11{{\left( \dfrac{-1}{3} \right)}^{3}}-55{{\left( \dfrac{-1}{3} \right)}^{2}}+163\left( \dfrac{-1}{3} \right)+60$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=3\times \dfrac{1}{81}-11\times \dfrac{-1}{27}-55\times \dfrac{1}{9}-\dfrac{163}{3}+60$
Simplifying further we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1}{27}+\dfrac{11}{27}-\dfrac{55}{9}-\dfrac{163}{3}+60$
Taking $27$ as LCM we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1+11-55\times 3-163\times 9+60\times 27}{27}$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1+11-165-1467+1620}{27}$
So we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{0}{27}$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=0$
As we get the value of polynomial zero at $x=-\dfrac{1}{3}$ so that means $3x+1$is the factor of the polynomial by the factor theorem.
Hence $3x+1$ is a factor of the polynomial $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-55{{x}^{2}}+163x+60$ .
Note: Factor theorem is usually used to determine the roots of the polynomial or to factorize the polynomial. We can also use the factor theorem to remove the known zeros from a polynomial while leaving the unknown zeros intact. Factor theorem is a special case of a polynomial remainder theorem. As we want the $x$ value we have to find the zero of the expression and that is the reason we have substituted $3x+1=0$ .
Complete step by step answer:
We have to determine whether $3x+1$ is a factor of the below polynomial:
$f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-55{{x}^{2}}+163x+60$….$\left( 1 \right)$
Now as we know that factor theorem states that if we have a polynomial $f\left( x \right)$ of degree$n\ge 1$ where $a$ is any real number than $\left( x-a \right)$ is a factor of the polynomial if the condition $f\left( a \right)=0$ is satisfied.
So as we have been given the expression as follows:
$3x+1$
$\Rightarrow 3x+1=0$
So we get,
$\Rightarrow 3x=-1$
$\Rightarrow x=-\dfrac{1}{3}$
So we get $x=-\dfrac{1}{3}$ substitute it in equation (1) as follows:
$\Rightarrow f\left( \dfrac{-1}{3} \right)=3{{\left( \dfrac{-1}{3} \right)}^{4}}-11{{\left( \dfrac{-1}{3} \right)}^{3}}-55{{\left( \dfrac{-1}{3} \right)}^{2}}+163\left( \dfrac{-1}{3} \right)+60$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=3\times \dfrac{1}{81}-11\times \dfrac{-1}{27}-55\times \dfrac{1}{9}-\dfrac{163}{3}+60$
Simplifying further we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1}{27}+\dfrac{11}{27}-\dfrac{55}{9}-\dfrac{163}{3}+60$
Taking $27$ as LCM we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1+11-55\times 3-163\times 9+60\times 27}{27}$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{1+11-165-1467+1620}{27}$
So we get,
$\Rightarrow f\left( \dfrac{-1}{3} \right)=\dfrac{0}{27}$
$\Rightarrow f\left( \dfrac{-1}{3} \right)=0$
As we get the value of polynomial zero at $x=-\dfrac{1}{3}$ so that means $3x+1$is the factor of the polynomial by the factor theorem.
Hence $3x+1$ is a factor of the polynomial $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-55{{x}^{2}}+163x+60$ .
Note: Factor theorem is usually used to determine the roots of the polynomial or to factorize the polynomial. We can also use the factor theorem to remove the known zeros from a polynomial while leaving the unknown zeros intact. Factor theorem is a special case of a polynomial remainder theorem. As we want the $x$ value we have to find the zero of the expression and that is the reason we have substituted $3x+1=0$ .
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