Solveeit Logo

Question

Question: How do you factor \({x^2} + 8x - 19\)?...

How do you factor x2+8x19{x^2} + 8x - 19?

Explanation

Solution

First compare the given quadratic equation to standard quadratic equation and find the value of numbers aa, bb and cc in given equation. Then, substitute the values of aa, bb and cc in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of aa, bb and DD in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity D=b24acD = {b^2} - 4ac is known as the discriminant of the equation ax2+bx+c=0a{x^2} + bx + c = 0 and its roots are given by
x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}} or x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step answer:
We know that an equation of the form ax2+bx+c=0a{x^2} + bx + c = 0, a,b,c,xRa,b,c,x \in R, is called a Real Quadratic Equation.
The numbers aa, bb and cc are called the coefficients of the equation.
The quantity D=b24acD = {b^2} - 4ac is known as the discriminant of the equation ax2+bx+c=0a{x^2} + bx + c = 0 and its roots are given by
x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}} or x=b±b24ac2ax = \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 aa, bb and cc.
Comparing x2+8x19{x^2} + 8x - 19 with ax2+bx+ca{x^2} + bx + c, we get
a=1a = 1, b=8b = 8 and c=19c = - 19
Now, substitute the values of aa, bb and cc in D=b24acD = {b^2} - 4ac and find the discriminant of the given equation.
D=(8)24(1)(19)D = {\left( 8 \right)^2} - 4\left( 1 \right)\left( { - 19} \right)
After simplifying the result, we get
D=140D = 140
Which means the given equation has real roots.
Now putting the values of aa, bb and DD in x=b±D2ax = \dfrac{{ - b \pm \sqrt D }}{{2a}}, we get
x=8±2352×1x = \dfrac{{ - 8 \pm 2\sqrt {35} }}{{2 \times 1}}
Divide numerator and denominator by 22, we get
x=4±35x = - 4 \pm \sqrt {35}
So, x=4+35x = - 4 + \sqrt {35} and x=435x = - 4 - \sqrt {35} are roots of equation x2+8x19=0{x^2} + 8x - 19 = 0.

Therefore,x2+8x19{x^2} + 8x - 19 can be factored as (x+435)(x+4+35)\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+35x = - 4 + \sqrt {35} and x=435x = - 4 - \sqrt {35} as the roots of equation x2+8x19=0{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.