How do you factor $6{{x}^{2}}-18x+12$ ?
Answer
558.3k+ views
Hint: First to factorize this given polynomial we must split the middle term. To split the middle term, we use the sum and product form. Now take-out common terms from the first two terms and last two terms. Then write them together as the product of sums form. Now represent them as the factors of the given expression.
Complete step by step solution:
The given polynomial which must be factorized is $6{{x}^{2}}-18x+12$
Let us first simplify the expression by taking 6 out as a common factor from all the terms.
$\Rightarrow 6\left( {{x}^{2}}-3x+2 \right)$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=1;b=-3;c=2$
The product of the terms is $2\times 1=2$
The sum of the terms is $-3$
Therefore, the terms can be $-2x,-x$
On substituting back, we get,
$\Rightarrow 6\left( {{x}^{2}}-2x-x+2 \right)$
The polynomial is of degree $2$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 6x\left( x-2 \right)-x+2$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 6x\left( x-2 \right)-\left( x-2 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow 6\left( x-2 \right)\left( x-1 \right)$
Now writing all the factors together we get,
$\Rightarrow 6\left( x-2 \right)\left( x-1 \right)$
Hence the factors for the polynomial $6{{x}^{2}}-18x+12$ are $6\left( x-2 \right)\left( x-1 \right)$
Note: The process of factorization is the reverse multiplication. In the above question, we have multiplied two linear line equations to get a quadratic equation (a polynomial of degree $2$) expression using the distributive law. The name quadratic comes from “quad” meaning square because the variable gets squared. Also known as a polynomial of degree $2$
Complete step by step solution:
The given polynomial which must be factorized is $6{{x}^{2}}-18x+12$
Let us first simplify the expression by taking 6 out as a common factor from all the terms.
$\Rightarrow 6\left( {{x}^{2}}-3x+2 \right)$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=1;b=-3;c=2$
The product of the terms is $2\times 1=2$
The sum of the terms is $-3$
Therefore, the terms can be $-2x,-x$
On substituting back, we get,
$\Rightarrow 6\left( {{x}^{2}}-2x-x+2 \right)$
The polynomial is of degree $2$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 6x\left( x-2 \right)-x+2$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 6x\left( x-2 \right)-\left( x-2 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow 6\left( x-2 \right)\left( x-1 \right)$
Now writing all the factors together we get,
$\Rightarrow 6\left( x-2 \right)\left( x-1 \right)$
Hence the factors for the polynomial $6{{x}^{2}}-18x+12$ are $6\left( x-2 \right)\left( x-1 \right)$
Note: The process of factorization is the reverse multiplication. In the above question, we have multiplied two linear line equations to get a quadratic equation (a polynomial of degree $2$) expression using the distributive law. The name quadratic comes from “quad” meaning square because the variable gets squared. Also known as a polynomial of degree $2$
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Find the sum of series 1 + 2 + 3 + 4 + 5 + + 100 class 9 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

What is pollution? How many types of pollution? Define it

Which are the Top 10 Largest States of India?

Name 10 Living and Non living things class 9 biology CBSE


