Question

How do you factor completely \[3{{x}^{2}}-9x-30\]?

Answer 1

Hint: The above expression has terms which have multiples of 3, we will begin by taking the common out. We will have the expression as, \[3({{x}^{2}}-3x-10)\]. We know that a quadratic equation is of the form \[{{x}^{2}}-(sum\text{ }of\text{ }zeroes)x+(product\text{ }of\text{ }zeroes)\]. And we have the factors as 5 and 2, we will use them appropriately and factor the given quadratic equation.

Complete step by step solution:
According to the given question, we are given a quadratic equation which we have to factorize completely.
We have the expression as,
\[3{{x}^{2}}-9x-30\]----(1)
We can see that the terms in the above expression has multiples of 3, so we can take it out as common factor, so we have,
\[3({{x}^{2}}-3x-10)\]----(2)
We will now factorize the expression \[{{x}^{2}}-3x-10\], we know that a quadratic equation is of the form \[{{x}^{2}}-(sum\text{ }of\text{ }zeroes)x+(product\text{ }of\text{ }zeroes)\].
We have the sum as 3 and the product as 10, we can think of the factors as 5 and 2, that is, we get,
\[\Rightarrow 3({{x}^{2}}-(5+(-2))x+5(-2))\]
Solving the above expression as
\[\Rightarrow 3({{x}^{2}}-(5-2)x+5(-2))\]
Opening up the brackets, we get the expression as,
\[\Rightarrow 3({{x}^{2}}-5x+2x+5(-2))\]
\[\Rightarrow 3({{x}^{2}}-5x+2x-10)\]
We will now take \[x\] common from the terms \[{{x}^{2}}\] and \[5x\]; and we will take 2 common from \[2x\] and \[10\], so we now have,
\[\Rightarrow 3(x(x-5)+2(x-5))\]
From the above expression, we will take \[(x-5)\]common and we have,
\[\Rightarrow 3((x-5)(x+2))\]
\[\Rightarrow 3(x-5)(x+2)\]
Therefore, the factors of the above expression is \[3(x-5)(x+2)\].

Note: The factors taken should fit perfectly with the standard form of a quadratic equation, which is, \[{{x}^{2}}-(sum\text{ }of\text{ }zeroes)x+(product\text{ }of\text{ }zeroes)\]. Also while taking the common factors out, the remaining terms should be written correctly, else will result in a wrong answer.