Question

How do you factor completely $2{x^2} - 98$ ?

Answer 1

To order to determine the factors of the above quadratic equation, first pull out $2$ common from the both the term and write the number $49$ as ${7^2}$ then use the identity $\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$ by considering $x$ as A and $7$ as B to factorise the expression completely.
Formula:
$\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$

Complete step by step solution:
Given a quadratic equation$2{x^2} - 98$,let it be $f(x)$
$f(x) = 2{x^2} - 98$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 2
b becomes 0
And c becomes -98
Lets Multiply and divide our expression with the number 2, we get
$f(x) = \dfrac{2}{2}(2{x^2} - 98) \\\ f(x) = 2\left( {\dfrac{{2{x^2}}}{2} - \dfrac{{98}}{2}} \right) \\\ f(x) = 2\left( {{x^2} - 49} \right) \\\$
As we know that $49$ can be written as $49 = 7 \times 7 = {7^2}$,we gte
$f(x) = 2\left( {{{\left( x \right)}^2} - {{\left( 7 \right)}^2}} \right)$
Consider $x$ as A and $7$ as B and Apply Identity $\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$
Now our equation becomes
$f\left( x \right) = \left( 2 \right)\left( {x - 7} \right)\left( {x + 7} \right)$
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are$\left( 2 \right)\left( {x - 7} \right)\left( {x + 7} \right)$.

Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.

Additional Information:
Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$ then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.