Completely factor the polynomial or state that the polynomial is prime.
w squared minus 29 w plus 168

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Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

Let's factor the given quadratic polynomial:

\[ w^2 - 29w + 168 \]

We need to find two numbers that multiply to 168 (the constant term) and add to -29 (the coefficient of the linear term).

The two numbers that satisfy these conditions are -21 and -8 because:

\[
-21 \times -8 = 168
\]
\[
-21 + (-8) = -29
\]

Thus, we can factor the quadratic as:

\[
w^2 - 29w + 168 = (w - 21)(w - 8)
\]

So the fully factored form of the polynomial is:

\[
\boxed{(w - 21)(w - 8)}
\]

This polynomial is not prime because it can be factored.

Would you like any more details or have any questions? Here are five related questions you might find useful:

1. How do you determine if a quadratic polynomial can be factored?
2. What is the relationship between the roots and the factors of a quadratic polynomial?
3. How can you verify that the factorization of a quadratic is correct?
4. What if the quadratic polynomial had a leading coefficient other than 1?
5. How would the process differ if the quadratic had complex roots?

**Tip:** When factoring quadratic polynomials, always check the discriminant \(\Delta = b^2 - 4ac\) first. If it's a perfect square, the quadratic can be factored using real numbers.

Learn how to completely factor the polynomial w^2 - 29w + 168 step-by-step. This problem involves factoring quadratic equations and understanding the relationship between roots and factors. Suitable for high school students.

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