To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (6, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (5, or b in the standard quadratic formula). Because their product is positive (6) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 2 and 3, as the product of 2 and 3 is 6, and sum of 2 and 3 is 5. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-9, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-9) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 3 and -3, as the product of 3 and -3 is -9, and sum of 3 and -3 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-3, or b in the standard quadratic formula). Because their product is positive (2) and the sum is negative, that must mean that they both have negative signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 2, as the product of 1 and 2 is 2, and sum of 1 and 2 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (2, or b in the standard quadratic formula). Because their product is positive (1) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 1, as the product of 1 and 1 is 1, and sum of 1 and 1 is 2. So, this results in the expression's factored form looking like...

From here, set the binomial equal to zero and solve for .

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zero of the function is,

To find the zeros of the function using factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-4, or b in the standard quadratic formula). Because their product is positive (4) and the sum is negative, that must mean that they both have negative signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are -2 and -2, as the product of -2 and -2 is 4, and sum of -2 and -2 is -4. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for . Since the binomials are the same, there will only be one zero.

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zero of the function is,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (3, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (4, or b in the standard quadratic formula). Because their product is positive (3) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and 3, as the product of 1 and 3 is 3, and sum of 1 and 3 is 4. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore, the zeros are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (20, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (9, or b in the standard quadratic formula). Because their product is positive (20) and the sum is positive, that must mean that they both have positive signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and 5, as the product of 4 and 5 is 20, and sum of 4 and 5 is 9. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-4, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (3, or b in the standard quadratic formula). Because their product is negative (-4) and the sum is positive, that must mean that they have opposite signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and -1, as the product of 4 and -1 is -4, and sum of 4 and -1 is 3. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

To find the zeros of the function use factoring. 

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-2, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (-1, or b in the standard quadratic formula). Because their product is negative (-2) and the sum is negative, that must mean that they have opposite signs. 

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -2, as the product of 1 and -1 is -2, and sum of -2 and 1 is -1. So, this results in the expression's factored form looking like...

From here, set each binomial equal to zero and solve for .

and 

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore, the zeros of the function are

To find the zeros of this function first identify and factor of the GCF.

In this particular case,the GCF is  as it appears in both terms. Factoring out the GCF results in the following.

From here, set each term equal to zero and solve for .

and

Therefore the zeros are,