Problem 85 - Elementary Algebra

Question

For exercises \(85-102\), factor completely. Identify any prime polynomials. $$ 2 p^{4}+4 p^{2} w^{3}+2 w^{6} $$

Step-by-Step Solution

Answer

2(p^{2} + w^{3})^{2}

Step 1: Identify Common Factors

First, check the given polynomial for any common factors in all the terms. The polynomial is 2p^{4}+4p^{2}w^{3}+2w^{6}. All the terms have a common factor of 2.

Step 2: Factor Out the Common Factor

Factor the common factor (2) out of each term in the polynomial: 2(p^{4}+2p^{2}w^{3}+w^{6}).

Step 3: Recognize the Quadratic Form

Notice that the polynomial inside the parentheses, p^{4}+2p^{2}w^{3}+w^{6}, looks like a quadratic in terms of p^{2}. Let's set u=p^{2}, so the polynomial becomes u^{2}+2uw^{3}+w^{6}. For comparison, this can be recognized as a square.

Step 4: Factor the Quadratic Form

Now, try to factor the quadratic form. u^{2}+2uw^{3}+w^{6} is a perfect square, which can be rewritten as (u + w^{3})^{2}. So the polynomial becomes: 2(p^{2} + w^{3})^{2}.

Step 5: Combine the Factors

Combine all the factors to get your final factored form: 2(p^{2} + w^{3})^{2}.

Key Concepts

common factors

When factoring polynomials, the first thing to look for are common factors. Common factors are numbers or variables that divide evenly into each term of the polynomial. Identifying common factors simplifies the polynomial, making it easier to factor completely.

In the given polynomial, $$2p^{4}+4p^{2}w^{3}+2w^{6},$$ each term contains the number 2 as a factor. You can factor out this 2, leading to:

$$2(p^{4} + 2p^{2}w^{3} + w^{6}).$$

Factoring out the greatest common factor reduces the complexity of the polynomial.

quadratic forms

After factoring out the common factor, it's important to look for patterns within the polynomial. One useful pattern is the quadratic form. Quadratic forms are expressions of the type $$ax^{2} + bx + c,$$ where the polynomial can be recognized as having similar characteristics to a regular quadratic equation.

In the polynomial

$$p^{4} + 2p^{2}w^{3} + w^{6},$$
we can substitute $$u = p^{2}$$ to recognize it as a quadratic in terms of $$u.$$ The polynomial then becomes

$$u^{2} + 2uw^{3} + w^{6}.$$

Identifying this form allows us to use our knowledge of quadratics to further factor the expression.

perfect squares

One of the easiest forms of quadratic to factor is a perfect square. A perfect square is a quadratic expression that can be written as the square of a binomial.

The quadratic:$$u^{2} + 2uw^{3} + w^{6},$$ is actually a perfect square. We can see that it fits the pattern $$(a + b)^{2} = a^{2} + 2ab + b^{2},$$ where $$a = u$$ and $$b = w^{3}.$$ Thus, it can be factored as $$(u + w^{3})^{2}.$$ This simplifies the original polynomial to

$$2(p^{2} + w^{3})^{2}.$$ Identifying and factoring out perfect squares simplifies the problem significantly.