Question

Factorise the equation \[{{x}^{3}}+13{{x}^{2}}+32x+20\]
(a). \[\left( x-1 \right)\left( x+2 \right)\left( x+5 \right)\]
(b). \[\left( x+1 \right)\left( x+2 \right)\left( x-5 \right)\]
(c). \[\left( x-2 \right)\left( x+3 \right)\left( x+10 \right)\]
(d). \[\left( x+1 \right)\left( x+2 \right)\left( x+10 \right)\]

Answer 1

Hint: To solve the question, we have to rearrange the terms of the expression to find common factors, the expression should be integrated into the sum of terms to ease the procedure of solving. Thus, we can narrow down the cubic equation to a quadratic equation and apply the same method of rearranging the terms to find the remaining two factors.

Complete step-by-step answer:

The given expression is \[{{x}^{3}}+13{{x}^{2}}+32x+20\]
The above expression can also be written as
\[\begin{align}
  & ={{x}^{3}}+{{x}^{2}}+12{{x}^{2}}+32x+20 \\
 & ={{x}^{3}}+{{x}^{2}}+12{{x}^{2}}+12x+20x+20 \\
\end{align}\]
By rearranging the terms of the equation, we get
\[={{x}^{2}}(x+1)+12x(x+1)+20(x+1)\]
By taking (x + 1) common in all the terms of the expression, we get
\[=(x+1)\left( {{x}^{2}}+12x+20 \right)\]
The above expression can also be written as
\[=(x+1)\left( {{x}^{2}}+10x+2x+20 \right)\]
\[=(x+1)\left( x(x+10)+2(x+10) \right)\]
By taking (x + 1) common in all the terms of the expression, we get
\[=(x+1)\left( \left( x+10 \right)\left( x+2 \right) \right)\]
\[=(x+1)\left( x+10 \right)\left( x+2 \right)\]
Thus, we get \[{{x}^{3}}+13{{x}^{2}}+32x+20=(x+1)\left( x+10 \right)\left( x+2 \right)\]
Hence, option (d) is the right answer.

Note: The possibility of mistake can be the calculation since the procedure of solving involves cubic equation solving. The alternative way of solving the question is by using, the formulae of general cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d\] and the formula for sum of the roots is given by \[\alpha +\beta +\gamma =\dfrac{-b}{a}\] and the formula for product of the roots is given by \[\alpha \beta \gamma =\dfrac{-d}{a}\] and the formula for sum of the product of roots taken two roots at a time is given by \[\alpha \beta +\beta \gamma +\alpha \gamma =\dfrac{c}{a}\] . Where \[\alpha ,\beta ,\gamma \] are the roots of the above equation. By substituting values in the above formula, we can calculate the factors of the equation. The other alternative way of solving the question is the option elimination method, since I know that for positive coefficients of an equation, the equation has only negative roots. Thus, we can eliminate the options that contain positive roots. Hence, we can arrive at the right choice.