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How do you factor $3{x^3} - 3{x^2} - 60x$

seo-qna
Last updated date: 17th Jun 2024
Total views: 373.8k
Views today: 9.73k
Answer
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373.8k+ views
Hint: In order to solve this question we will first put this equation equals to 0. After this, we will take all the commons and after it, the equation will be converted into the quadratic form then we will find the roots of that quadratic and after the transformation, we can find the factors.

Complete step by step solution:
For solving this we need to take all the possible commons:
$3{x^3} - 3{x^2} - 60x$
On taking all commons
$3x({x^2} - x - 20)$
Since one factor is already found the other we will find through the quadratic equations;
By putting the whole quadratic equation equals to 0.
${x^2} - x - 20 = 0$
On applying the dharacharya on this equation:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
From the above equation putting the value of a=1, b=-1, c=-20 on this formula:
$x = 5, - 4$
Now transforming these on side of x the factors will be:
$(x - 5)$ and $(x + 4)$
Now putting all the factors together with:

$(3x)(x - 5)(x + 4)$
So this will be the final answer.


Note:
This can also be found through an alternate method:
After taking the common we will direct factorize the quadratic equation;
$3x({x^2} - x - 20)$
Now factoring the quadratic:
$3x({x^2} - 5x + 4x - 20)$
Now taking all the possible commons:
$3x\{ x(x - 5) + 4(x - 5)\} $
Now further taking the common:
$(3x)(x - 5)(x + 4)$
So by this method also we find the same as the factors.
All the time we cannot apply the factorization method because we cannot find the factor when there are no real roots so the first one is more comfortable for us.