Question

How do you factor $ 4 - 25{x^2} $ ?

Answer 1

Hint: To order to determine the factors of the above quadratic equation, Rewrite the $ 4\,as\,{2^2} $ and $ 25\,as\,{5^2} $ then use the identity $ \left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right) $ by considering $ 2 $ as A and $ 5x $ as B to obtain the complete factorisation of the given expression.

Formula Used:
 $ \left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right) $

Complete step-by-step solution:
Given a quadratic equation $ 4 - 25{x^2} $ ,let it be $ f(x) $
 $ f(x) = 4 - 25{x^2} $
Comparing the equation with the standard Quadratic equation $ a{x^2} + bx + c $
a becomes -25
b becomes 0
And c becomes 4
As you can see $ 25 $ can be written as $ 25 = 5 \times 5 = {5^2} $ and similarly $ 4 = 2 \times 2 = {2^2} $ , so rewriting these values in the expression , we get
 $ f(x) = {\left( 2 \right)^2} - {\left( {5x} \right)^2} $
Now Consider $ 2 $ as A and $ 5x $ as B and Apply Identity $ \left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right) $
Now our expression becomes
 $ f\left( x \right) = \left( {2 - 5x} \right)\left( {2 + 5x} \right) $
Hence, we have successfully factorized our quadratic equation.
Therefore, the factors are $ \left( {2 - 5x} \right)\left( {2 + 5x} \right) $ .

Note:
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 linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $ D = {b^2} - 4ac $
i) Using Discriminant, we can find out the nature of the roots
ii) If D is equal to zero, then both of the roots will be the same and real.
iii) If D is a positive number then, both of the roots are real solutions.
iv) If D is a negative number, then the root are the pair of complex solutions