Polynomials - Math

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is . To put this in expanded form:

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is , which we can expland using the cube of a binomial pattern.

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

Then, we are looking for the values where each of these factors are equal to zero.

implies

and implies

Thus, these are our solutions.

First, multiply the outside term with each term within the parentheses:

Rearranging the polynomial into fractional form, we get:

To simplify the polynomial, begin by combining like terms:

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

Since the function equals zero when is , one of the factors of the polynomial is . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

Now we can factor the function this way:

(x%5E%7B3%7D-5x%5E%7B2%7D-2x+24) $"f(x)=(x+1)(x^{3}$$-5x^{2}$-2x+24)")

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

When we factor using synthetic substitution for , we get the following result:

Using our quadratic factoring rules, we can factor completely:

Thus, the zeroes of are

First, multiply the outside term with each term within the parentheses:

Rearranging the polynomial into fractional form, we get:

To simplify the polynomial, begin by combining like terms:

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

We are able to get back to the original expression, meaning that the answer is .

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is . To put this in expanded form:

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is , which we can expland using the cube of a binomial pattern.

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

Then, we are looking for the values where each of these factors are equal to zero.

implies

and implies

Thus, these are our solutions.

First, we need to find all the possible rational roots of the polynomial using the Rational Roots Theorem:

Since the leading coefficient is just 1, we have the following possible (rational) roots to try:

±1, ±2, ±3, ±4, ±6, ±12, ±24

When we substitute one of these numbers for , we're hoping that the equation ends up equaling zero. Let's see if is a zero:

Since the function equals zero when is , one of the factors of the polynomial is . This doesn't help us find the other factors, however. We can use synthetic substitution as a shorter way than long division to factor the equation.

Now we can factor the function this way:

(x%5E%7B3%7D-5x%5E%7B2%7D-2x+24) $"f(x)=(x+1)(x^{3}$$-5x^{2}$-2x+24)")

We repeat this process, using the Rational Roots Theorem with the second term to find a possible zero. Let's try :

When we factor using synthetic substitution for , we get the following result:

Using our quadratic factoring rules, we can factor completely:

Thus, the zeroes of are

Knowing the zeroes makes it relatively easy to factor the polynomial.

The expression fits the description of the zeroes.

Now we need to check the answer.

We are able to get back to the original expression, meaning that the answer is .

Since 5 is a single root and 7 is a double root, and the degree of the polynomial is 3, the polynomial is . To put this in expanded form:

Since 6 is a triple root, and the degree of the polynomial is 3, the polynomial is , which we can expland using the cube of a binomial pattern.

When we are looking for the solutions of a quadratic, or the zeroes, we are looking for the values of such that the output will be zero. Thus, we first factor the equation.

Then, we are looking for the values where each of these factors are equal to zero.

implies

and implies

Thus, these are our solutions.

First, multiply the outside term with each term within the parentheses:

Rearranging the polynomial into fractional form, we get:

To simplify the polynomial, begin by combining like terms: