Question

How do you solve ${x^3} - 5{x^2} - 14x = 0$

Answer 1

Hint: In the question, we have an equation that is quadratic, and we have to find its factors, which can be done with the help of rules. Factorisation is the reverse of multiplying out. One important factorisation process is the reverse of multiplications. The factors of any equation can be an integer, a variable or an algebraic expression itself.
Here we can take out the common multiplier and simplify the equation in order to factorise it by finding roots.

Complete step-by-step solution:
We have an equation ${x^3} - 5{x^2} - 14x = 0$, and we are supposed to find the value of the variable which can only be done after finding its roots with the help of factorisation. First step is to factor out $x$ from each term on the left side of the equation to leave $x$ and a quadratic equation:
Taking out the common multiple, the equation now looks like,
\[ \Rightarrow x(x2 - 5x - 14) = 0\]
Next, factor the quadratic equation:
\[ \Rightarrow x(x - 7)(x + 2) = 0\]
Next solve for each term $x$, \[(x - 7)\] and \[(x + 2)\] to obtain the three solutions for this problem.
1) Solve for $x$:
\[ \Rightarrow x\dfrac{{(x - 7)(x + 2)}}{{(x - 7)(x + 2)}} = \dfrac{0}{{(x - 7)(x + 2)}}\]
Therefore, we get \[x = 0\]
2) Solve for \[(x - 7)\]:
\[ \Rightarrow x\dfrac{{(x - 7)(x + 2)}}{{x(x + 2)}} = \dfrac{0}{{x(x + 2)}}\]
Solving the equation,
\[ \Rightarrow x - 7 = 0\]
Therefore, the final value is,
\[ \Rightarrow x = 7\]
3) Solve for \[(x + 2)\]:
\[ \Rightarrow x\dfrac{{(x - 7)(x + 2)}}{{x(x - 7)}} = \dfrac{0}{{x(x - 7)}}\]
Solving the equation,
\[ \Rightarrow x + 2 = 0\]
Therefore, the final value is,
\[ \Rightarrow x = - 2\]

Therefore, for an equation ${x^3} - 5{x^2} - 14x = 0$, the value of the variable will be $0,7, - 2$ as per the situation.

Note: In Mathematics, factorisation or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together gives the original number or a matrix, etc. It is simply the resolution of an integer or polynomial into factors such that when multiplied together they will result in an original or initial integer or polynomial. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets.