Question

How do you factor \[3{x^2} - 12x\]?

Answer 1

Hint: Here in this question, we have to find the factors, the given equation is in the form of a quadratic equation. This is a quadratic equation for the variable x. By using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we can determine the roots of the equation and factors are given by (x – root1) (x – root 2)

Complete step-by-step solution:
The question involves the quadratic equation. To the quadratic equation we can find the roots by factoring or by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. So the equation is written as \[3{x^2} - 12x\].

In general, the quadratic equation is represented as \[a{x^2} + bx + c = 0\], when we compare the above equation to the general form of equation the values are as follows. a=3 b=-12 and c=0. Now substituting these values to the formula for obtaining the roots we have
\[roots = \dfrac{{ - ( - 12) \pm \sqrt {{{( - 12)}^2} - 4(3)(0)} }}{{2(3)}}\]
On simplifying the terms, we have
\[ \Rightarrow roots = \dfrac{{12 \pm \sqrt {144 - 0} }}{6}\]
Now add 144 to 0 we get
\[ \Rightarrow roots = \dfrac{{12 \pm \sqrt {144} }}{6}\]
The number 144 is a perfect square so we can take out from square root we have
\[ \Rightarrow roots = \dfrac{{12 \pm 12}}{6}\]
Therefore, we have \[root1 = \dfrac{{12 + 12}}{6} = \dfrac{{24}}{6} = 4\] or \[root2 = \dfrac{{12 - 12}}{6} = 0\].
The roots for the quadratic equation when we find the roots by using formula is given by (x – root1) (x – root 2)
Substituting the roots values, we have
\[ \Rightarrow \left( {x - 4} \right)\left( {x - 0} \right)\]
On simplifying we have
\[ \Rightarrow x(x - 4)\]
This can also be solved by another method.
Consider the given equation \[3{x^2} - 12x\]
Divide the equation by 3 we get
\[{x^2} - 4x\]
Take x as a common so we have
\[x(x - 4)\]

Hence, we have found the factors for the given equation

Note: The quadratic equation can be solved by using the factorisation method and we also find the roots by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. While factorising we use sum product rule, the sum product rule is given as the product factors of the number c is equal to the sum of the factors which satisfies the value of b.