Finding Zeros - SAT Math

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 (-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 (0, or b in the standard quadratic formula). Because their product is negative (-1) 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 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 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 (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 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 (16, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (8, or b in the standard quadratic formula). Because their product is positive (16) 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 4, as the product of 4 and 4 is 16, and sum of 4 and 4 is 8. 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 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,

The important first step of creating a quadratic equation from its zeros is knowing what a zero really is. A zero of a quadratic (or polynomial) is an x-coordinate at which the y-coordinate is equal to 0.

We find that algebraically by factoring quadratics into the form , and then setting equal to and , because in each of those cases and entire parenthetical term would equal 0, and anything times 0 equals 0.

Here the question gives you a head start: we know that the numbers 4 and 5 can go in the and spots, because if so we'll have found our zeros. So we can set up the equation:

This satisfies the requirements of zeros, but now we need to expand this equation using FOIL to turn it into a proper quadratic. That means that our quadratic is:

And when we combine like terms it's:

The zeros of a function are points at which the function crosses the x-axis, or perhaps more simply points at which the function is equal to 0. So to solve for those, set the function equal to zero and then solve it like you would a quadratic. Here that gives you:

You can then factor the common term:

And then factor the quadratic within parentheses. Note that this quadratic is one of the common perfect square quadratics:

or

The solutions to this equation, then, are and . Note that the question asks you for how many distinct zeros the function has, so you cannot count twice. The answer, then, is 2.

A function with zeros at 2 and 7 would factor to , where it is important to recognize that is the coefficient. So while you might be looking to simply expand to , note that none of the options with a simple term (and not ) directly equal that simple quadratic when set to 0.

However, if you multiply by 2, you get .

When you're graphing a function, the zeros of the function are the points at which the function crosses the x-axis. What that really means is that the value of the function is zero at those points, so to solve for those zeros algebraically you can just set the function equal to zero and solve. Here that would mean:

So factor the common term to get:

And then factor the quadratic within parentheses:

You can then see that there are three values of that would make this equation true: and . The answer is therefore 3.

The zeros of a function are the x-values at which the function is equal to zero. So to solve for the zeros, set the function equal to zero. That would give you here:

Then you can factor the common to get:

And then factor like you would a quadratic:

Or, more succinctly formatted:

This means that the zeros are at and . Now, importantly, look at what the question asks for. It wants the sum (add the zeros) of all UNIQUE zeros, meaning you should not count twice. The sum then is .

The zeros of a function are points at which the function is equal to zero. Here you're told that does not equal zero for any real number values of and you're asked to determine what that means for . Note that since cannot be negative, would have to be negative in order for the term to reduce the other term, to zero. Since you know that this function never equals zero, you can conclude that is not negative, which means that it is greater than or equal to zero.

The zeros of a function are the x-values at which the function itself is equal to zero, so you will generally want to solve these problems by setting the function equal to zero and then factoring like a quadratic. Here that means you would start with:

And then factor the common term:

You now know that one of the solutions is , as that would mean that the entire parenthetical term would be multiplied by zero. But you still need to work within the parentheses to factor that quadratic. You should see some helpful factoring clues: when you factor into two parentheticals, the numeric terms have to multiply to -1, meaning that you will likely have one be 1 and the other -1. And the first terms need to multiply to meaning that your first terms will be and . So you can set up your parentheses as:

, where one will have a + and one will have a - sign.If you play with the options to see what will work to give you the middle term in , you'll see that the proper factorization is:

This means that the solutions for are and . The sum of these solutions, then, is .

The zeros of a function are essentially points (or at least the x-values of the points) at which the function is equal to zero. So to solve for the zeros of a function, first set that function itself equal to zero. Here that would mean:

Then factor like you would a quadratic; since you have it set to zero, if any multiplicative term equals zero then the "equals zero" will hold for the whole equation. First you can factor the common term:

And then you can factor the quadratic within:

This then means that the zeros for this function are at and , meaning that this function has three distinct zeros.