Finding Zeros of a Polynomial - Math

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:

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, 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:

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, 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:

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, 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:

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