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Question: How do you factor completely \(36{x^2} - 25\) ?...

How do you factor completely 36x2−2536{x^2} - 25 ?

Explanation

Solution

To order to determine the factors of the above quadratic equation using the identity (A2−B2)=(A−B)(A+B)\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)

Complete step by step solution:
Given a quadratic equation 36x2−2536{x^2} - 25,let it be f(x)f(x)
f(x)=36x2−25f(x) = 36{x^2} - 25
Comparing the equation with the standard Quadratic equation
a becomes 36
b becomes 0
And c becomes -25
To find the quadratic factorization we’ll be writing the expression as
f(x)=(6x)2−(5)2f\left( x \right) = {(6x)^2} - {(5)^2}
Consider 6x6xas A and 55as B and Applying Identity (A2−B2)=(A−B)(A+B)\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)
Now our equation becomes
f(x)=(6x−5)(6x+5)f\left( x \right) = (6x - 5)(6x + 5)
Hence, We have successfully factorized our quadratic equation.
Therefore, the factors are(6x−5)(6x - 5) and(6x+5)(6x + 5)

Alternative:
You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
x1=−b+b2−4ac2ax1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} and x2=−b−b2−4ac2ax2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}
x1,x2 are root to quadratic equation ax2+bx+ca{x^2} + bx + c
Hence the factors will be (x−x1) and (x−x2) (x - x1)\,and\,(x - x2)\,.

Additional Information:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of ax2+bx+ca{x^2} + bx + c where xx is the unknown variable and a,b,c are the numbers known where a≠0a \ne 0. If a=0a = 0then 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=b2−4acD = {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.