How do you factor\[2{x^2} - x - 15?\]

Hint:The given question describes the operation of addition/ subtraction/ multiplication/ division. Also, remind the basic form of a quadratic equation and quadratic formula which is used to find the value of\[x\]. And, compare the given equation with the quadratic formula to solve the question. We need to know the root value of basic numbers.

Complete step by step solution:
The given equation is shown below,
\[2{x^2} - x - 15 = 0 \to \left( 1 \right)\]

We know that the quadratic formula is,
\[a{x^2} + bx + c = 0 \to \left( 2 \right)\]

Then, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]

Let’s compare the equation\[\left( 1 \right)\]and\[\left( 2 \right)\], for finding the values
of\[a,b\]and\[c\].

\[\left( 1 \right) \to 2{x^2} - x - 15 = 0\]
\[\left( 2 \right) \to a{x^2} + bx + c = 0\]

So, let’s compare the\[{x^2}\]terms in the equation\[\left( 1 \right)\]and\[\left( 2 \right)\]
\[
2 \times {x^2} \\
a \times {x^2} \\
\]

So, we find\[a = 2\].

Let’s compare the\[x\]terms in the equation\[\left( 1 \right)\]and\[\left( 2 \right)\]
\[
- 1 \times x \\
b \times x \\
\]

So, we find\[b = - 1\].

Let’s compare the constant terms in the equation\[\left( 1 \right)\]and\[\left( 2 \right)\]

So, we find\[c = - 15\].

So, we get\[a,b\]and\[c\]values are\[2, - 1\]and\[ - 15\]respectively.

Let’s substitute this value in the equation\[\left( 3 \right)\]for finding the value of\[x\]

\[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 2 \times - 15} }}{{2
\times 2}}\]

By solving the above equation we get,
\[
x = \dfrac{{1 \pm \sqrt {1 + 120} }}{4} \\
x = \dfrac{{1 \pm \sqrt {121} }}{4} \\
\]

We know that\[\sqrt {121} = 11\], so the above equation becomes,
\[x = \dfrac{{1 \pm 11}}{4}\]

Due to the presence of\[ \pm \]we get two values for\[x\].

Case: 1
\[x = \dfrac{{1 + 11}}{4}\]
\[x = \dfrac{{12}}{4}\]
\[x = 3\]

Case: 2
\[
x = \dfrac{{1 - 11}}{4} \\
x = \dfrac{{ - 10}}{4} \\
x = \dfrac{{ - 5}}{2} \\
\]
In case: 1 we assume\[ \pm \]as an\[ + \]operation, in case: 2 we assume\[ \pm \]as an\[ - \]
operation.

By considering the\[ \pm \]as\[ + \]we get \[x = 3\] and by considering the\[ \pm \]as\[ -
\]we get\[x = \dfrac{{ - 5}}{2}\].

So, the final answer is, \[x = 3\]or\[x = \dfrac{{ - 5}}{2}\].

Note: To find the value of\[x\] from the given equation we would compare the equation with the quadratic formula. After comparing the equations we would find the value of\[a,b\]and\[c\]. When substituting these values in quadratic formula remind the following things,

1) When a negative number is multiplied with a negative number, the answer becomes a
positive number.

2) When a positive number is multiplied with a positive number, the answer becomes a
positive number.

3) When a negative number is multiplied with a positive number, the answer becomes
negative.