Factors of a Quadratic Polynomial

We know that any quadratic polynomial can be written as $f(x)=ax^2+bx+c$ , where a, b and c are real numbers. We have following method to factor a quadratic polynomial. 

First of all check whether c is positive or negative. 

If c is positive. 

Step 1: Find the product of a and c. To get product of a and c just multiply a with c.

Step 2: Now get all the factors of the product ac. 

Step 3: Find two factors of the product ac such that sum of the two factors is equal to  the b i.e. the coefficient of x.
Step 4: Plug in that sum in place of b and simplify to get the factors. 

Let us take an example. Consider $p(x)=2x^2+5x+3$
Compare this polynomial with $f(x)=ax^2+bx+c$, we get $a=2, b=5 \, and \, c=3$
Step 1: Product of  a and c. $2\times 3=6$
Step 2: All the factors of ac i.e. 6 are 1, 2, 3, 6
Step 3: We have to find two factors of 6 such that their sum is equal to b i.e. 5
If we consider factors 2 and 3, we have $2+3=5$ and $2\times 3=6$
Step 4: We substitute this sum in place of b and simplify to get the factors.
            We have  $p(x)=2x^2+(2+3)x+3$
                                     $=2x^2+2x+3x+3$
                                     $=2x(x+1)+3(x+1)$
                                     $=(x+1)(2x+3)$

This is how we get factors of a quadratic polynomial when c is positive. 

If c is negative then we follow the following step

Step 1: Find the product of a and c. To get product of a and c just multiply a with c.

Step 2: Now write all the factors of the product ac. 

Step 3: Find two factors of the product ac such that difference of the two factors is equal to  the b i.e. the coefficient of x.
Step 4: Plug in that difference in place of b and simplify and get the factors. 

Let us take an example. Consider $p(x)=2x^2+x-3$
Compare this polynomial with $f(x)=ax^2+bx+c$, we get $a=2, b=1 \, and \, c=-3$
Step 1: Product of  a and c. $2\times 3=6$
Step 2: All the factors of ac i.e. 6 are 1, 2, 3, 6
Step 3: We have to find two factors of 6 such that their difference is equal to b i.e. 1
If we consider factors 2 and 3, we have $3-2=1$ and $2\times 3=6$

Step 4: We substitute this difference in place of b and simplify and get the factors.

            We have  $p(x)=2x^2+(3-2)x-3$
                                     $=2x^2+3x-2x-3$
                                     $=x(2x+3)-(2x+3)$
                                     $=(2x+3)(x-1)$

This is how we get factors of a quadratic polynomial when c is negative.   

Besides it, we can use box method, or graphic method to find the factors of a quadratic polynomial.
If we put p(x)=0, we get two values for x which satisfy $p(x)=0$. These two values of x are known as the zeros of the given Quadratic polynomial. Let's take the our first one example $p(x)=2x^2+5x+3$
Here, the factored form of p(x) is $p(x)=(x+1)(2x+3)$ 
And $p(x)=0$ gives us $(x+1)(2x+3)=0$ 
Here, the product of two factors (x+1) and (2x+3) is zero. The product can be zero if either of two is zero. Hence, either $x+1=0$  or $2x+3=0$
Or we can say either $x=-1$  or $x=-\frac{3}{2}$
Hence, zeros of $p(x)=2x^2+5x+3$ are $-1$ and  $-\frac{3}{2}$

Similarly, we can find the zeros of any quadratic polynomial. 

Find the factors of given problems 
$1.  \, 4x^2+14x-8$
$2. \,  3x^2+10x+7$
Find the zeros of the polynomial
$1. \, p(x)=3x^2-11x+8$
$2. \, q(x)=6x^2+9x-6$