
Using factor theorem show that \[g\left( x \right)\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\], \[g\left( x \right)=x+3\].
Answer
410.1k+ views
Hint: In order to show that \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\], we must solve for \[g\left( x \right)\] at first. Then we are supposed to substitute this value of \[g\left( x \right)\] in \[p\left( x \right)\] and solve the polynomial. If the value of \[p\left( x \right)\] would be zero, then the given \[g\left( x \right)\] will be a factor of \[p\left( x \right)\].
Complete step-by-step solution:
Now let us learn about the factor theorem. The factor theorem is a theorem that links the factors and zeros of a polynomial. This is a special case of the remainder theorem. According to this theorem we say that \[x-a\] is a factor of \[f\left( x \right)\], if \[f\left( a \right)=0\]. This theorem is commonly used to find the roots of the polynomial. We can also find the factor of a polynomial by other methods such as the polynomial long division method and the synthetic division method.
Now let us find if \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
Firstly, we must be solving for \[g\left( x \right)=x+3\].
We get,
\[\begin{align}
& g\left( x \right)=x+3 \\
& \Rightarrow x+3=0 \\
& \Rightarrow x=-3 \\
\end{align}\]
We obtain the value of \[x\] as \[-3\].
Now we will be substituting this obtained value in \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
\[\begin{align}
& \Rightarrow p\left( x \right)=69+11x-{{x}^{2}}+{{x}^{3}} \\
& \Rightarrow p\left( -3 \right)=69+11\left( -3 \right)-{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{3}} \\
& \Rightarrow p\left( -3 \right)=69-33-9-27 \\
& \Rightarrow p\left( -3 \right)=0 \\
\end{align}\]
We see that we have obtained \[p\left( x \right)=0\].
\[\therefore \] We can conclude that given \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
Note: To apply the factor theorem, we must always have a note that for a polynomial \[f\left( x \right)\] the degree of the polynomial should be greater than or equal to one. The degree of the polynomial is nothing but the highest power of the term in the expression. The value which solves the expression or equation is known as polynomial value.
Complete step-by-step solution:
Now let us learn about the factor theorem. The factor theorem is a theorem that links the factors and zeros of a polynomial. This is a special case of the remainder theorem. According to this theorem we say that \[x-a\] is a factor of \[f\left( x \right)\], if \[f\left( a \right)=0\]. This theorem is commonly used to find the roots of the polynomial. We can also find the factor of a polynomial by other methods such as the polynomial long division method and the synthetic division method.
Now let us find if \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
Firstly, we must be solving for \[g\left( x \right)=x+3\].
We get,
\[\begin{align}
& g\left( x \right)=x+3 \\
& \Rightarrow x+3=0 \\
& \Rightarrow x=-3 \\
\end{align}\]
We obtain the value of \[x\] as \[-3\].
Now we will be substituting this obtained value in \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
\[\begin{align}
& \Rightarrow p\left( x \right)=69+11x-{{x}^{2}}+{{x}^{3}} \\
& \Rightarrow p\left( -3 \right)=69+11\left( -3 \right)-{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{3}} \\
& \Rightarrow p\left( -3 \right)=69-33-9-27 \\
& \Rightarrow p\left( -3 \right)=0 \\
\end{align}\]
We see that we have obtained \[p\left( x \right)=0\].
\[\therefore \] We can conclude that given \[g\left( x \right)=x+3\] is a factor of \[p\left( x \right)\]\[=69+11x-{{x}^{2}}+{{x}^{3}}\].
Note: To apply the factor theorem, we must always have a note that for a polynomial \[f\left( x \right)\] the degree of the polynomial should be greater than or equal to one. The degree of the polynomial is nothing but the highest power of the term in the expression. The value which solves the expression or equation is known as polynomial value.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
