Question

How do you factor \[3{{x}^{2}}+6x-9\]?

Answer 1

Hint: In this problem, we have to find the factors of the given quadratic equation. We can take the common term outside the given quadratic equation. Then we will get a simplified quadratic equation, from which we can take the constant term, which is the result of multiplying two numbers and when adding/subtracting the same two numbers, we will get the coefficient of x, those two numbers are the factors of the equation.

Complete step by step answer:
We know that the given quadratic equation to be factored is,
 \[3{{x}^{2}}+6x-9\]
We can take the common number outside the equation, we get
\[\Rightarrow 3\left( {{x}^{2}}+2x-3 \right)\] …… (2)
Now we can factorize the new quadratic equation \[\left( {{x}^{2}}+2x-3 \right)\].
We can take the constant term -3, if we multiply 3 and -1 we will get the constant term -3, if we add 3 and -1, we will get the coefficient of x, i.e., 2.
\[\Rightarrow 3\times \left( -1 \right)=-3\], the constant term.
\[\Rightarrow 3+\left( -1 \right)=2\], the coefficient of x.
The factors are \[\left( x-3 \right)\left( x+1 \right)\] .
We can substitute the factors in the equation (1), we get
\[\Rightarrow 3\left( x-3 \right)\left( x+1 \right)\]
Therefore, the factors of the equation \[3{{x}^{2}}+6x-9\] are \[3\left( x-3 \right)\left( x+1 \right)\].

Note:
Students make mistakes while taking common terms form the given equation, which should be concentrated. We can also use a perfect square method to factorize the given equation or we can split the middle term into parts and take common terms from that to factorize the given equation. The number, which is taken as a common number outside in the first step should be multiplied with the factors to get the correct answer.