Polynomials - 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:

(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.

Factor:

Double check by factoring:

Add together:

Therefore:

1. Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

1. Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

1. Now pull out the common factor, the "(x-2)," from both terms.

1. Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

To solve for , you need to isolate it to one side of the equation. You can subtract the from the right to the left. Then you can add the 6 from the right to the left:

Next, you can factor out this quadratic equation to solve for . You need to determine which factors of 8 add up to negative 6:

Finally, you set each binomial equal to 0 and solve for :

If we recognize this as an expression with form , with and , we can solve this equation by factoring:

and

and

To find the values for in which the polynomial equals , we first want to factor the equation:

When finding zeros of a polynomial, you must remember your rules. Without a function this may seem tricky, but remember that non-real solutions come in conjugate pairs. Conjugate pairs differ in the middle sign. Thus, our answer is: