Question

How do you factor ${x^2} + 8x - 19$?

Answer 1

Hint: First compare the given quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in 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 answer:
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 compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + 8x - 19$ with $a{x^2} + bx + c$, we get
$a = 1$, $b = 8$ and $c = - 19$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 8 \right)^2} - 4\left( 1 \right)\left( { - 19} \right)$
After simplifying the result, we get
$D = 140$
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
$x = \dfrac{{ - 8 \pm 2\sqrt {35} }}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$x = - 4 \pm \sqrt {35} $
So, $x = - 4 + \sqrt {35} $ and $x = - 4 - \sqrt {35} $ are roots of equation ${x^2} + 8x - 19 = 0$.

Therefore,${x^2} + 8x - 19$ can be factored as $\left( {x + 4 - \sqrt {35} } \right)\left( {x + 4 + \sqrt {35} } \right)$.

Note: In above question, it should be noted that we get $x = - 4 + \sqrt {35} $ and $x = - 4 - \sqrt {35} $ as the roots of equation ${x^2} + 8x - 19 = 0$. 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.