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How do you factor by grouping ${{x}^{3}}+3{{x}^{2}}+6x=18?$

seo-qna
Last updated date: 16th Jun 2024
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Answer
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Hint: (1) The term factor by grouping then we have to factor out common factors. Then we have to compare the given equation in the standard form, i.e. $a{{x}^{2}}+bx+c$
(2) Then we have to substitute the given equation in standard form. But here the equation started with the cubic term. So, you need to do this by grouping the first two terms with each other and the second term with each other.
(3) After that take out the common factor from it you get the factor by the factor by grouping.

Complete step by step solution:Here we have,
Here,
We have the equation,
${{x}^{3}}+3{{x}^{2}}+6x+18$
The equation shows that the ratio between the first and second terms are the same as that between the third and fourth term.
Then we can say that the method we use here is factor by grouping to solve problems.
${{x}^{3}}+3{{x}^{2}}+6x+18$
Now,
[Grouping the first two terms with each other and second term with each other]
We get,
${{x}^{3}}+3{{x}^{2}}+6x+18=\left( {{x}^{3}}+3{{x}^{2}} \right)+\left( 6x+18 \right)$ [common factor]
Taking common from brackets.
${{x}^{3}}+3{{x}^{2}}+6x+18={{x}^{2}}\left( x+3 \right)+6\left( x+3 \right)$
$=\left( {{x}^{2}}+6 \right)\left( x+3 \right)$
Hence,
By factor by grouping we get the factor $\left( {{x}^{2}}+6 \right)\left( x+3 \right)$

Additional Information:
(1) Factoring in a polynomial has writing it as a product of two or more polynomials.
(2) In the method of factorisation we reduce it in simplex form. The factor can be an integer, a variable or algebraic itself in any type of equation.
(3) There are various methods of factorisation. They are factoring out GCF, the grouping method, the difference of square pattern etc.
(4) We have several example of factoring, However for this you should taking common factor using distributive property.
For example: $6{{x}^{2}}+4x=2x\left( 3x+2 \right)$

Note:
(1) To use a grouping method we have to group the expression in two groups. As we have two groups factor out the GCF from the given groups.
(2) If you solve this type of equation the binomial which is in the parenthesis must be the same, if it is not the same then there should be any mistake in the factor.
(3) you should also check the sign properly while writing the final answer.