Question

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

Answer 1

Hint: To order to determine the factors of the above quadratic equation, Rewrite the $4as{2}^2$ and $25as{5}^2$ 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 $4-25x^2$ ,let it be $f(x)$
 $f(x)=4-25x^2$
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