Question

How do you factor completely : $ 5{x^2} + 20x $ ?

Answer 1

Hint: To simplify and then factor the equation completely as per this question , we need to solve it step by step . The first thing we should always do while factoring is find a common factor. In this case, it would be 5x as it divides into both of these . Then take each variable and divide it by your factor. Then just put your factor on the outside of your bracket and the two other variables inside and you've got your answer .

Complete step by step solution:
Usually what we do to factorize the equation and to determine the factors of the above quadratic question use the Splitting up the middle method by first multiplying the coefficient of with the constant term and factorise it into two factors such that either addition or subtraction gives us the middle term and the product of the same gives us back the multiplication we have calculated. Pull out common from the first two terms and the last two terms and then again pulling out the binomial parenthesis will give you the required factorisation .
But according to our question simple quadratic equation is given to us $ 5{x^2} + 20x $
Now we are going to find a common factor from the given equation which is going to be $ 5x $ as it is the only common number which divides into both of the terms .
After taking common and to represent mathematically , then take each variable and divide it by your factor by this we will get –
\[
  \dfrac{{5{x^2}}}{{5x}} = x \\
  \dfrac{{20x}}{{5x}} = 4 \;
 \]
Now we will just put our factor $ 5x $ outside the bracket and the other two variables we got inside the bracket representing it mathematically as = $ 5x(x + 4) $ which is our required answer .
So, the correct answer is “ $ 5x(x + 4) $ ”.

Note: In algebra, a quadratic equation is any equation that can be rearranged in standard form as $ ax^{2}+bx+c=0 $ where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no $ ax^2 $ term.