Question

How do you find the factors of \[5{{x}^{3}}+15{{x}^{2}}-20x\]?

Answer 1

Hint: This type of question is based on the concept of factorisation. We can solve this question by first taking the common terms out of the bracket. Here 5x is the common term. Then, make some necessary calculations and take the other common terms outside the bracket by splitting the middle terms. Thus, we get the factors of the given equation \[5{{x}^{3}}+15{{x}^{2}}-20x\].

Complete answer:
According to the question, we are asked to find the factors of the given equation \[5{{x}^{3}}+15{{x}^{2}}-20x\].
We have been given the equation is \[5{{x}^{3}}+15{{x}^{2}}-20x\]. ------(1)
Now, we observe that 5x is common throughout the given equation (1).
Therefore, let us take 5x common.
\[5{{x}^{3}}+15{{x}^{2}}-20x=5x\left( {{x}^{2}}+3x-4 \right)\] ----------(2)
From equation (2) we find that 5x is a factor of \[5{{x}^{3}}+15{{x}^{2}}-20x\].
Now let us consider \[{{x}^{2}}+3x-4 \] -------(3)
Let us first split the middle term of the equation (3).
We know that 3x=4x-x.
Substituting this in equation (3), we get,
\[{{x}^{2}}+3x-4={{x}^{2}}+4x-x-4\]
\[\Rightarrow {{x}^{2}}+3x-4={{x}^{2}}-x+4x-4\]
Now, let us take x common from the first two terms and -4 common from the last two terms.
\[{{x}^{2}}+3x-4=x\left( x-1 \right)+4\left( x-1 \right)\] ---------(4)
 Let us take \[\left( x-1 \right)\] common from equation (4).
\[\Rightarrow {{x}^{2}}+3x-4=\left( x-1 \right)\left( x+4 \right)\]
Therefore, the factors are \[5x\], \[\left( x-1 \right)\] and \[\left( x+4 \right)\].
Hence, the factors of \[5{{x}^{3}}+15{{x}^{2}}-20x\] are \[5x\], \[\left( x-1 \right)\] and \[\left( x+4 \right)\].

Note: Whenever you get this type of problem, we should always try to make the necessary calculations in the given equation to get the final of x. We should then find the factors as required. We should also avoid calculation mistakes based on sign conventions. Similarly, we can find the factors of quadratic as well as cubic equations.