Question

How do you factor completely ${{x}^{3}}-5{{x}^{2}}+6x$?

Answer 1

Hint: Now to solve the given expression we will first take x common from the whole expression. Now we will factorize the obtained quadratic by splitting the middle terms. Hence we will simplify the expression and find the factors of the quadratic expression. Hence we will have the factors of the given expression.

Complete step by step solution:
Now we are given the expression ${{x}^{3}}-5{{x}^{2}}+6x$.
Note that the expression has no constant term. Hence we can say that 0 is the root of the expression.
Now since 0 is the root of the expression we have $\left( x-0 \right)$ as a factor of the expression.
Now taking x common from the whole expression we get,
$\Rightarrow x\left( {{x}^{2}}-5x+6 \right)$
Now we will consider the quadratic ${{x}^{2}}-5x+6$ .
Now to factorize the quadratic expression we will split the middle terms which is $-5x$
Now we will split the middle term such that the product of the terms is multiplication of the first term and last term.
Hence we have, $-5x=-3x-2x$ since $\left( -3x \right)\left( -2x \right)=6{{x}^{2}}$
Hence we get the quadratic expression as ${{x}^{2}}-3x-2x+6$ .
Now taking x common from the first two terms and -2 common from the last two terms we get,
$\begin{align}
  & \Rightarrow x\left( x-3 \right)-2\left( x-3 \right) \\
 & \Rightarrow \left( x-2 \right)\left( x-3 \right) \\
\end{align}$
Hence we get the factorization of the quadratic is $\left( x-2 \right)\left( x-3 \right)$ .
Hence substituting this value we get the factorization of the given expression as $x\left( x-2 \right)\left( x-3 \right)$ .
Hence the factors of the given expression are x, x-2 and x-3.

Note: Now note that if we have no constant in the expression we always take x common to simplify the expression. Also note that while splitting the middle term we cannot write $-5x=-6x+x$ as the product of the terms is $-6{{x}^{2}}$ and the product of the first term and the last term is $6{{x}^{2}}$ .