Write a Polynomial Function from its Zeros - Algebra 2

The degree is the highest exponent value of the variables in the polynomial.

Here, the highest exponent is x5, so the degree is 5.

By the Rational Zeroes Theorem, any rational solution must be a factor of the constant, 6, divided by the factor of the leading coefficient, 14.

Four of the answer choices have this characteristic:

is in lowest terms, and 3 is not a factor of 14. It is therefore the correct answer.

from the zeroes given and the Fundamental Theorem of Algebra we know:

use FOIL method to obtain:

Distribute:

Simplify:

Find a polynomial function of the lowest order possible such that two of the roots of the function are:

Recall that by roots of a polynomial we are referring to values of such that .

Because one of the roots given is a complex number, we know there must be a second root that is the complex conjugate of the given root. This is .

Because is a root, the unknown function must have a factor

The other roots are complex numbers, so there must be a quadratic factor.

To find the quadratic factor start with the value for one of the complex roots:

Isolate the imaginary term onto one side and square,

Expand the left side, and note on the right side the factor reduces as follows:

So now we have,

The quadratic factor for is therefore . Combining this with the factor give a factored expression for the desired function:

Now we carry out the multiplication to write the final form of ,

A quadratic function takes the form where x is a variable and a, b, and c are coefficients. Equations of this nature can be factored, and then each factor can be set equal to zero and then solved to find the roots of the equation. This problem asks you to take that idea, but work through it in reverse. To solve it, begin with:

and

Get everyone on one side of the equation:

and

These are the factors of the equation. Put them together, and then multiply them to get:

Therefore, the solution to this question is

A quadratic 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 . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .

When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:

So you know that a possible quadratic for these zeros would be:

Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in , the result of the quadratic will be -24. So you can set up the equation:

, where is the coefficient you're solving for. And you know that this is true when so if you plug in 2, you can solve for :

This means that:

So:

And therefore . When you then distribute that coefficient of 3 across the entire quadratic:

So:

When you're finding the equation of a quadratic given its zeros, a good start is to ignore the coefficient (you'll return to it later) and to construct the factored form of the quadratic using the zeros. Here with zeros of 5 and 2, that factored form would be:

That then means you can expand it to:

From here, you can see that the answer choices all maintain the same general relationship as this simple quadratic, but have different coefficients. This is where you can use the point (1, 16) to your advantage. This means that if you plug in to the quadratic, it will produce an answer of . To solve this algebraically, return to using a coefficient in front of your entire quadratic. You'll then have:

Where is the coefficient you can solve for knowing that the equation will produce a value of when you input . So plug that in and you can solve for the coefficient:

Meaning that:

So you know that .

Then you can multiply that coefficient across your simple quadratic to get the answer:

Becomes:

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 answer is .

A quadratic with zeros at x = 2 and x = 7 would factor to , where it is important to recognize that a 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 . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .

When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:

So you know that a possible quadratic for these zeros would be:

Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in x = 2, the result of the quadratic will be -24. So you can set up the equation:

, where a is the coefficient you’re solving for. And you know that this is true when x = 2, so if you plug in x = 2 you can solve for a:

So a = 3. When you then distribute that coefficient of 3 across the original simple quadratic, you have:

So your quadratic is

When you're writing a quadratic having been given its zeros, the best place to start is by setting aside the coefficient and first putting together a simple, factored quadratic that would satisfy those zeros. Here with zeros at 7 and -3, that would be:

If you then expand that quadratic, you have:

Of course, that's just one possible quadratic that satisfies those zeros, and your job is to find THE quadratic that satisfies those zeroes AND passes through (0, -63). And clearly here if you plug in to the quadratic you would not get . So your next step is to determine the coefficient. You can do that by setting the value of the coefficient as :

And then plugging in since you know that when .

So

Meaning that . You can then multiply the original quadratic by 3 to get:

This can be written as:

Multiply the terms together:

Multiply the first two terms:

FOIL:

Combine like terms:

The degree is the highest exponent value of the variables in the polynomial.

Here, the highest exponent is x5, so the degree is 5.

By the Rational Zeroes Theorem, any rational solution must be a factor of the constant, 6, divided by the factor of the leading coefficient, 14.

Four of the answer choices have this characteristic:

is in lowest terms, and 3 is not a factor of 14. It is therefore the correct answer.

from the zeroes given and the Fundamental Theorem of Algebra we know:

use FOIL method to obtain:

Distribute:

Simplify:

Find a polynomial function of the lowest order possible such that two of the roots of the function are:

Recall that by roots of a polynomial we are referring to values of such that .

Because one of the roots given is a complex number, we know there must be a second root that is the complex conjugate of the given root. This is .

Because is a root, the unknown function must have a factor

The other roots are complex numbers, so there must be a quadratic factor.

To find the quadratic factor start with the value for one of the complex roots:

Isolate the imaginary term onto one side and square,

Expand the left side, and note on the right side the factor reduces as follows:

So now we have,

The quadratic factor for is therefore . Combining this with the factor give a factored expression for the desired function:

Now we carry out the multiplication to write the final form of ,

A quadratic function takes the form where x is a variable and a, b, and c are coefficients. Equations of this nature can be factored, and then each factor can be set equal to zero and then solved to find the roots of the equation. This problem asks you to take that idea, but work through it in reverse. To solve it, begin with:

and

Get everyone on one side of the equation:

and

These are the factors of the equation. Put them together, and then multiply them to get:

Therefore, the solution to this question is

A quadratic 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 . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .