Answer
Verified
430.8k+ views
Hint: We have been given a polynomial expression $p\left( x \right)$ and also given another expression $x - a$ is a factor of the given polynomial. We have to determine the remaining factors of the given polynomial. According to the remainder theorem, if $x - a$ is a factor of $p\left( x \right)$ then $p\left( a \right) = 0$ . To find the other factors of the given polynomial, first, we divide the given polynomial $p\left( x \right)$ by $x - a$ and determine the quotient. After that, we factorize the quotient, using middle term splitting. The obtained factors are the other two factors of the given polynomial.
Complete step by step answer:
Given a polynomial is $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ and a factor is $x + 9$. We have to determine the other two factors of the above given polynomial. For that, first we perform the long division of the given polynomial $f\left( x \right)$ by the factor $x + 9$, we get
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} - 7x + 12 $
$ x + 9)\overline {{x^3} + 2{x^2} - 51x + 108} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\underline {{x^3} + 9{x^2}} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 7{x^2} - 51x $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, { - 16{x^2} - 63x} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{12x + 108 }$
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {12x + 108} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 $
The quotient after dividing the given polynomial $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ by $x + 9$ , we get ${x^2} - 7x + 12$ . Now we factorize this quadratic equation using middle term splitting.
First, we split the middle term $ - 7x$ as $ - 4x - 3x$ because the addition of both is equal to the middle term and the product is equal to the 1product of the first and last term of the quadratic equation. So, after middle term splitting, we get
$ \Rightarrow {x^2} - 4x - 3x + 12$
Now we take common from first two terms and last two terms, we get
$ \Rightarrow x\left( {x - 4} \right) - 3\left( {x - 4} \right)$
Now take $\left( {x - 4} \right)$ common, we get
$ \Rightarrow \left( {x - 4} \right)\left( {x - 3} \right)$
So the other two factors of the given polynomials are $\left( {x - 4} \right)$ and $\left( {x - 3} \right)$ .
Note: In middle term splitting, split the middle term such that the addition of the split terms is equal to the middle term and the product of the split terms is equal to the product of the first and last term of the quadratic equation.
Complete step by step answer:
Given a polynomial is $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ and a factor is $x + 9$. We have to determine the other two factors of the above given polynomial. For that, first we perform the long division of the given polynomial $f\left( x \right)$ by the factor $x + 9$, we get
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} - 7x + 12 $
$ x + 9)\overline {{x^3} + 2{x^2} - 51x + 108} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\underline {{x^3} + 9{x^2}} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 7{x^2} - 51x $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, { - 16{x^2} - 63x} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline{12x + 108 }$
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {12x + 108} $
$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 $
The quotient after dividing the given polynomial $f\left( x \right) = {x^3} + 2{x^2} - 51x + 108$ by $x + 9$ , we get ${x^2} - 7x + 12$ . Now we factorize this quadratic equation using middle term splitting.
First, we split the middle term $ - 7x$ as $ - 4x - 3x$ because the addition of both is equal to the middle term and the product is equal to the 1product of the first and last term of the quadratic equation. So, after middle term splitting, we get
$ \Rightarrow {x^2} - 4x - 3x + 12$
Now we take common from first two terms and last two terms, we get
$ \Rightarrow x\left( {x - 4} \right) - 3\left( {x - 4} \right)$
Now take $\left( {x - 4} \right)$ common, we get
$ \Rightarrow \left( {x - 4} \right)\left( {x - 3} \right)$
So the other two factors of the given polynomials are $\left( {x - 4} \right)$ and $\left( {x - 3} \right)$ .
Note: In middle term splitting, split the middle term such that the addition of the split terms is equal to the middle term and the product of the split terms is equal to the product of the first and last term of the quadratic equation.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE