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Question

Question: How do you factor \({x^3} - 25x?\)...

How do you factor x325x?{x^3} - 25x?

Explanation

Solution

First check if there any common factor between the terms of the given expression, if yes then take the common factor out and rewrite the expression after taking the common factor and check if there any algebraic identity could be applied or not, if yes then apply that. The algebraic identity a2b2=(a+b)(ab){a^2} - {b^2} = (a + b)(a - b) will be used to solve this question.

Complete step by step solution:
In order to factorize the expression x325x{x^3} - 25x, we need first to check for any common factor between the terms of the given expression,

Let us check the common factor between x3{x^3} and 25x25x by listing their factors as follows

x3=x×x×x 25x=5×5×x  {x^3} = x \times x \times x \\\ 25x = 5 \times 5 \times x \\\
We can see that xx is the common factor between x3{x^3} and 25x25x
So taking out xx common in the expression x325x{x^3} - 25x, we will get
=x325x =x(x225)  = {x^3} - 25x \\\ = x({x^2} - 25) \\\
We got the new expression =x(x225) = x({x^2} - 25), but this is not the factored form of x325x{x^3} - 25x
Check the expression x(x225)x({x^2} - 25), can we factorize it further?
Yes, we can factorize it furthermore by use of an algebraic identity which could be given as
a2b2=(a+b)(ab){a^2} - {b^2} = (a + b)(a - b)

See the (x225)({x^2} - 25) in the expression x(x225)x({x^2} - 25), we can see that (x225)({x^2} - 25) is
suitable for the above algebraic identity, so factoring further with the help of a2b2=(a+b)(ab){a^2} - {b^2} = (a + b)(a - b), we will get =x(x225) = x({x^2} - 25)

We can write 25=5225 = {5^2}
=x(x252) =x((x+5)(x5)) =x(x+5)(x5)  = x({x^2} - {5^2}) \\\ = x((x + 5)(x - 5)) \\\ = x(x + 5)(x - 5) \\\
Therefore x(x+5)(x5)x(x + 5)(x - 5) is the required factored form of x325x{x^3} - 25x

Note: In order to factorize any polynomial expression, checking for common factors is very helpful way that is if you get any common factor then after taking out the common factor the expression becomes more simplified and then for further factoring, other methods will be easily visible to you after taking out the common factor.