Find all real zeros of f(x) = 2x^2 +5x-18 algebraically.

2x2-5x-18=0

Two solutions were found :

x = -2
x = 9/2 = 4.500
Step by step solution :

Step 1 :

Equation at the end of step 1 :

(2x2 - 5x) - 18 = 0
Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring 2x2-5x-18

The first term is, 2x2 its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -18

Step-1 : Multiply the coefficient of the first term by the constant 2 • -18 = -36

Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .

-36 + 1 = -35
-18 + 2 = -16
-12 + 3 = -9
-9 + 4 = -5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4
2x2 - 9x + 4x - 18

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (2x-9)
Add up the last 2 terms, pulling out common factors :
2 • (2x-9)
Step-5 : Add up the four terms of step 4 :
(x+2) • (2x-9)
Which is the desired factorization

Equation at the end of step 2 :

(2x - 9) • (x + 2) = 0
Step 3 :

Theory - Roots of a product :

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.