Question

How do you factor the expression \[{x^2} - 5x - 14\]?

Answer 1

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we need to know the basic form of a quadratic equation and the formula to find the value \[x\]from the quadratic equation. We need to know the basic square root and square values of basic numbers. We have the term \[{x^2}\]in the given question, so we would find two values \[x\] by solving the given equation.

Complete step-by-step solution:
The given equation is shown below,
\[{x^2} - 5x - 14\]
The above equation can also be written as,
\[y = {x^2} - 5x - 14 \to \left( 1 \right)\]
We know that the basic form of a quadratic equation is,
\[a{x^2} + bx + c = y \to \left( 2 \right)\]
Then the formula for finding the value of\[x\]from the above equation is shown below,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
By comparing the equation\[\left( 1 \right)\& \left( 2 \right)\], we get the value of\[a,b\]and\[c\]
\[\left( 1 \right) \to y = {x^2} - 5x - 14\]
\[\left( 2 \right) \to a{x^2} + bx + c = y\]
So, we get the value of\[a\]is\[1\], the value of\[b\]is\[ - 5\] , and the value of\[c\]is\[ - 14\].
Let’s substitute these values in the equation\[\left( 3 \right)\], we get
\[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[
  x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4 \times 1 \times - 14} }}{{2 \times 1}} = \dfrac{{5 \pm \sqrt {25 + 56} }}{2} = \dfrac{{5 \pm \sqrt {81} }}{2} \\
  x = \dfrac{{5 \pm 9}}{2} \\
 \]
Let’s solve the above equation in two cases as follows,
Case: \[1\]
\[
  x = \dfrac{{5 \pm 9}}{2} \Rightarrow x = \dfrac{{5 + 9}}{2} = \dfrac{{14}}{2} \\
  x = 7 \\
 \]
Case: \[2\]
\[
  x = \dfrac{{5 \pm 9}}{2} \Rightarrow x = \dfrac{{5 - 9}}{2} = \dfrac{{ - 4}}{2} \\
  x = - 2 \\
 \]

So, the final answer is \[x = 7,x = - 2\]Or\[\left( {x - 7} \right),\left( {x + 2} \right)\]

Note: This question describes the arithmetic operations like addition/ subtraction/ multiplication/ division. Note that the denominator term would not be equal to zero. When \[{n^2}\]is placed inside the square root, we can cancel the square root and square each other. It\[ \pm \]is present in the equation we would find two values for\[x\]. When multiplying different sign terms, we would remember the following things,
i) When a negative term is multiplied/ divided with the negative term the final answer would be a positive term.
ii) When a positive term is multiplied/ divided with the positive term the final answer would be a positive term.
iii) When a positive term is multiplied/ divided with the negative term the final answer would be a negative term.