Finding Zeros of a Polynomial - College Algebra

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

Now we factor out the (3x + 4).

Setting each factor = 0 we can find the solutions.

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

By the Rational Zeroes Theorem, any rational zeroes of a polynomial must be obtainable by dividing a factor of the constant coefficient by a factor of the leading coefficient. Since both values are equal to 1, and 1 has only 1 as a factor, this restricts the set of possible rational zeroes to the set .

Both values can be tested as follows:

1 is a zero of if and only if . An easy test for this is to add the coefficients and determine whether their sum, which is , is 0:

1 is indeed a zero.

is a zero of if and only if . An easy test for this is to add the coefficients after changing the sign of the odd-degree coefficients, and determine whether their sum, which is , is 0. However, as their are no odd-degree coefficients, the sum is the same:

is also a zero.

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: