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Chapter 0: Problem 113

Factor completely. $$2 x^{2}-7 x y^{2}+3 y^{4}$$

Short Answer

Expert verified
The expression \(2 x^{2}-7 x y^{2}+3 y^{4}\) cannot be factored further since it鈥檚 a prime polynomial.

Step by step solution

01

Identifying the Form

Identify the form of the expression. In this case, the expression \(2 x^{2}-7 x y^{2}+3 y^{4}\) is a trinomial, a polynomial having three terms.
02

Identifying Possible Factor Pairs

Find pairs of factors of the first and last term whose sum or difference provide the coefficient of the middle term. The factor pairs of \(2x^{2}\) are \((2x, x)\) and the factor of \(3y^{4}\) is \((y^{2},3y^{2})\).
03

Factoring

Under normal circumstances, we would try to identify the factors that meet above conditions. But the factor pairs here do not result in a value that is equivalent to the coefficient of the middle term which is \(-7xy^{2}\). Thus, the trinomial cannot be factored using integer coefficients. The trinomial is a prime polynomial, so it cannot be factored further.

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Most popular questions from this chapter

Fill in each box to make the statement true. $$ \sqrt{x}=5 x^{7} $$

Factor Completely. $$y^{7}+y$$

What is a perfect square trinomial and how is it factored?

In parts (a) and (b), complete each statement. $$\text { a. } b^{4} \cdot b^{3}=(b \cdot b \cdot b \cdot b)(b \cdot b \cdot b)=b^{?}$$ $$\text { b. } b^{5} \cdot b^{5}=(b \cdot b \cdot b \cdot b \cdot b)(b \cdot b \cdot b \cdot b \cdot b)=b^{7}$$ c. Generalizing from parts (a) and (b), what should be done with the exponents when multiplying exponential expressions with the same base?

How do the whole numbers differ from the natural numbers?
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