Question

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

Answer 1

Hint: This question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the basic form of the quadratic equation and formula to find the value of \[x\] in a quadratic equation. We need to know the basic square root values of basic numbers. We have the term \[x\] square in the question. So, we would find two answers \[x\] by solving the given equation.

Complete step-by-step answer:
The given question is shown below,
 \[3{x^2} - x - 2 = 0 \to \left( 1 \right)\]
We know that the basic form of a quadratic equation is,
 \[a{x^2} + bx + c = 0 \to \left( 2 \right)\]
The formula for finding the value \[x\] from the above equation is shown below,
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
By comparing the equation \[\left( 1 \right)\] and \[\left( 2 \right)\] , we get the values of \[a,b\] and \[c\] .
 \[\left( 1 \right) \to 3{x^2} - x - 2 = 0\]
 \[\left( 2 \right) \to a{x^2} + bx + c = 0\]
So, we get the value of \[a\] is \[3\] , the value of \[b\] is \[ - 1\] , and the value of \[c\] is \[ - 2\] . Let’s substitute these value in the equation \[\left( 3 \right)\] , we get
 \[\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
 \[x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \times 3 \times - 2} }}{{2 \times 3}}\]
 \[
  x = \dfrac{{1 \pm \sqrt {1 + 24} }}{6} \\
  x = \dfrac{{1 \pm \sqrt {25} }}{6} \;
 \]
We know that \[{5^2} = 25\] . So, the above equation can also be written as,
 \[x = \dfrac{{1 \pm \sqrt {{5^2}} }}{6}\]
So, we get
 \[x = \dfrac{{1 \pm 5}}{6}\]
Case: 1
 \[
  x = \dfrac{{1 + 5}}{6} \\
  x = \dfrac{6}{6} \\
  x = 1 \;
 \]
Case: 2
 \[
  x = \dfrac{{1 - 5}}{6} \\
  x = \dfrac{{ - 4}}{6} \\
  x = \dfrac{{ - 2}}{3} \;
 \]
So, the final answer is,
 \[x = 1\] and \[x = \dfrac{{ - 2}}{3}\]
So, the correct answer is “ \[x = 1\] and \[x = \dfrac{{ - 2}}{3}\] ”.

Note: This type of question involves the arithmetic operation of addition/ subtraction/ multiplication/ division. Note that the denominator term would not be equal to zero. Note that zero divided by anything becomes zero and anything divided by zero becomes infinity. When \[{n^2}\] is placed inside the square root, we can cancel the square and the square root can be cancelled by each other. If \[ \pm \] is present in the equation we would find two values for \[x\] . If \[{x^2}\] is present in the equation we will have two answers for \[x\] by solving the given equation.