Question

How do you factor $ - 3{x^2} + 2x + 2$?

Answer 1

Hint: First take $ - 3$ common from the given equation. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used: The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
It is given that the expression of $ - 3{x^2} + 2x + 2$
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will take $ - 3$ common from the given equation.
$ - 3{x^2} + 2x + 2 = - 3\left( {{x^2} - \dfrac{2}{3}x - \dfrac{2}{3}} \right)$
Next, compare ${x^2} - \dfrac{2}{3}x - \dfrac{2}{3}$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - \dfrac{2}{3}x - \dfrac{2}{3}$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - \dfrac{2}{3}$ and $c = - \dfrac{2}{3}$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - \dfrac{2}{3}} \right)^2} - 4\left( 1 \right)\left( { - \dfrac{2}{3}} \right)$
After simplifying the result, we get
$D = \dfrac{{28}}{9}$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{\dfrac{2}{3} \pm \dfrac{{2\sqrt 7 }}{3}}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$x = \dfrac{1}{3} \pm \dfrac{{\sqrt 7 }}{3}$
So, $x = \dfrac{1}{3} + \dfrac{{\sqrt 7 }}{3}$ and $x = \dfrac{1}{3} - \dfrac{{\sqrt 7 }}{3}$are roots of equation ${x^2} - \dfrac{2}{3}x - \dfrac{2}{3}$.

Therefore, the trinomial $ - 3{x^2} + 2x + 2$ can be factored as $ - 3\left( {x - \dfrac{1}{3} - \dfrac{{\sqrt 7 }}{3}} \right)\left( {x - \dfrac{1}{3} + \dfrac{{\sqrt 7 }}{3}} \right)$.

Note: In above question, it should be noted that we get $x = \dfrac{1}{3} + \dfrac{{\sqrt 7 }}{3}$and $x = \dfrac{1}{3} - \dfrac{{\sqrt 7 }}{3}$ as the roots of equation${x^2} - \dfrac{2}{3}x - \dfrac{2}{3}$.
No other roots will satisfy the condition.
If we take wrong factors, then we will not get a trinomial on their product. So, carefully find the roots.