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Question

Question: How do you factor \( {x^2} + 7x + 12 \) ?...

How do you factor x2+7x+12{x^2} + 7x + 12 ?

Explanation

Solution

In this question, we need to factor x2+7x+12{x^2} + 7x + 12 .
First, we will find two numbers that multiply to give acac i.e., 1×12=121 \times 12 = 12 , and add to give bb i.e., 77 which is called a sum-product pattern. Then rewrite the middle with those numbers. Then, factor the first two and last two terms separately. If we have done this correctly, then two new terms will have a clearly visible common factor.

Complete step-by-step solution:
Now, we need to factor x2+7x+12{x^2} + 7x + 12 .
According to the rule to factorize,
Product= x2{x^2} coefficient ×\times constant
And, sum= xx coefficient
Thus, we will find two numbers that multiply to give acac i.e., 1×12=121 \times 12 = 12 and add to give bb
i.e., 77 ,
Here, the product is negative. So, we can say that one of the factors is negative, and then the other is positive.
Now, let’s consider the possible factors and their sum.
12×1=12;12+1=1312 \times 1 = 12;12 + 1 = 13
6×2=12;6+2=86 \times 2 = 12;6 + 2 = 8
4×3=12;4+3=74 \times 3 = 12;4 + 3 = 7
From this it is clear that the factors are 33 and 44 .
Now, by rewriting the middle term with those factors, we have,
=x2+3x+4x+12= {x^2} + 3x + 4x + 12
Factor out the greatest common factor from each group,
=x(x+3)+4(x+3)= x\left( {x + 3} \right) + 4\left( {x + 3} \right)
Factor the polynomial by factoring out the greatest common factor, x+3x + 3 ,
= (x+4)(x+3)\left( {x + 4} \right)\left( {x + 3} \right)

Hence, the factors of x2+7x+12{x^2} + 7x + 12 is (x+4)\left( {x + 4} \right) and (x+3)\left( {x + 3} \right) .

Note: In this question it is important to note here that this factorization method works for all quadratic equations. The standard form of the quadratic equation is ax2+bx+c=0a{x^2} + bx + c = 0 . It is called factoring because we find the factors. A factor is something we multiply by. There is no simple method of factoring a quadratic expression, but with a little practice it becomes easier. Here, the middle step can be skipped with some practice.