Question

How do you factor\[{{x}^{3}}+{{x}^{2}}-5x+3\]?

Answer 1

Hint: In order to solve this question, firstly find a root \[x=a\]which fits the equation \[{{x}^{3}}+{{x}^{2}}-5x+3=0\]perfectly. Then, just divide the factor \[\left( x-a \right)\]with the equation that we need to factorize and we get a quadratic equation as the remainder. The quadratic equation is factored then.

Formula used: The formula used in here is that for the first factor, we simply put the value of a root and solve the equation to get a zero. If we get a zero then, \[\left( x-a \right)\]is a factor for the equation.
Now after that, we divide the equation with the factor \[\left( x-a \right)\] to get a quotient that gives the second factor, that quotient is a quadratic equation that is further split into factors.

Complete step-by-step solution:
Let us start solving the question by taking the equation,
\[{{x}^{3}}+{{x}^{2}}-5x+3=0\]
Now, let us think of a root that completely satisfies this equation,
If
 \[\begin{align}
  & x=1 \\
 & \Rightarrow {{1}^{3}}+{{1}^{2}}-5\left( 1 \right)+3=2+3-5=0 \\
\end{align}\]
So, we can say that the factor \[\left( x-1 \right)\]satisfies the equation, now dividing the equation \[{{x}^{3}}+{{x}^{2}}-5x+3\]with\[\left( x-1 \right)\], we get
\[\left( x-1 \right){{\left| \!{\overline {\,
 \begin{align}
  & {{x}^{3}}+{{x}^{2}}-5x+3 \\
 & \underline{_{-}{{x}^{3}}{{-}_{+}}{{x}^{2}}}\downarrow +\downarrow \\
 & 2{{x}^{2}}-5x \\
 &\underline{ _{-}2{{x}^{2}}{{-}_{+}}2x} \\
 & -3x+3 \\
 & -3x+3 \\
 & =0 \\
\end{align} \,}} \right. }^{{{x}^{2}}+2x-3}}\]
Therefore, we can say that
\[\left( x-1 \right)\left( {{x}^{2}}+2x-3 \right)={{x}^{3}}+{{x}^{2}}-5x+3\]
So, now, we just need to make the further factors for the factor \[\left( {{x}^{2}}+2x-3 \right)\]
\[\begin{align}
  & \left( {{x}^{2}}+2x-3 \right) \\
 & \Rightarrow {{x}^{2}}-x+3x-3 \\
 & \Rightarrow x\left( x-1 \right)+3\left( x-1 \right) \\
 & \Rightarrow \left( x+3 \right)\left( x-1 \right) \\
\end{align}\]
Thus, we can write that the factors for the equation\[{{x}^{3}}+{{x}^{2}}-5x+3\] are
\[\begin{align}
  & \left( x+3 \right)\left( x-1 \right)\left( x-1 \right) \\
 & \Rightarrow \left( x+3 \right){{\left( x-1 \right)}^{2}} \\
\end{align}\]
Hence the equation has been factored.

Note: The given equation is a cubic equation, that is why we get three factors. However, as the factor \[\left( x-1 \right)\]is repeated twice, its square has been taken but the degree of the equation still remains the same i.e. \[3\]. The factors can be verified by further multiplying them to get the same equation as before.