Find Complex Zeros of a Polynomial Using the Fundamental Theorem of Algebra - Pre-Calculus

One of the roots is because if we plug in 1, we get 0. We can factor the polynomial as

So now we solve the roots of .

The root will not be real.

The roots of this polynomial are .

So, the roots are

If this polynomial has a real zero at 1.5, that means that the polynomial has a factor that when set equal to zero has a solution of . We can figure out what this is this way:

multiply both sides by 2

is the factor

Now that we have one factor, we can divide to find the other two solutions:

To finish solving, we can use the quadratic formula with the resulting quadratic, :

Since we know that one of the zeros of this polynomial is 3, we know that one of the factors is . To find the other two zeros, we can divide the original polynomial by , either with long division or with synthetic division:

This gives us the second factor of . We can get our solutions by using the quadratic formula:

We know that the real zero of this polynomial is 3, so one of the factors must be . To find the other factors, we can divide the original polynomial by , either by long division or synthetic division:

This gives us a second factor of which we can solve using the quadratic formula:

First, factorize the equation using grouping of common terms:

Next, setting each expression in parenthesis equal to zero yields the answers.

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has solutions at two values: when and when

Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations:

which are then set equal to zero to yield the expressions and

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

First, factor the expression by grouping:

To find the complex zeroes, set the term equal to zero:

First, factorize the equation using grouping of common terms:

Next, setting each expression in parentheses equal to zero yields the answers.

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has a solution as two values: when , and when . Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations: which are then set equal to zero to yield the expressions and

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

First, factor the expression by grouping:

To find the complex zeroes, set the term equal to zero:

One of the roots is because if we plug in 1, we get 0. We can factor the polynomial as

So now we solve the roots of .

The root will not be real.

The roots of this polynomial are .

So, the roots are

If this polynomial has a real zero at 1.5, that means that the polynomial has a factor that when set equal to zero has a solution of . We can figure out what this is this way:

multiply both sides by 2

is the factor

Now that we have one factor, we can divide to find the other two solutions:

To finish solving, we can use the quadratic formula with the resulting quadratic, :

Since we know that one of the zeros of this polynomial is 3, we know that one of the factors is . To find the other two zeros, we can divide the original polynomial by , either with long division or with synthetic division:

This gives us the second factor of . We can get our solutions by using the quadratic formula:

We know that the real zero of this polynomial is 3, so one of the factors must be . To find the other factors, we can divide the original polynomial by , either by long division or synthetic division:

This gives us a second factor of which we can solve using the quadratic formula:

First, factorize the equation using grouping of common terms:

Next, setting each expression in parenthesis equal to zero yields the answers.

First, pull out the common t and then factorize using quadratic factoring rules:

This equation has solutions at two values: when and when

Therefore, But since the degree on the former equation is one and the degree on the latter equation is two, the multiplicities are 1 and 2 respectively.

This one is a bit of a journey. The expressions for the first two zeroes are easily calculated, and respectively. The last expression must be broken up into two equations:

which are then set equal to zero to yield the expressions and

Finally, we multiply together all of the parenthesized expressions, which multiplies out to

First, factor the expression by grouping:

To find the complex zeroes, set the term equal to zero: