Question

1. factor the expression completely: $10x^5 - 2x$

Explanation:

Step1: Find the GCF of coefficients and variables

The GCF of 10 and 2 is 2. For the variable part, the GCF of \(x^5\) and \(x\) is \(x\). So the GCF of \(10x^5\) and \(2x\) is \(2x\).

Step2: Factor out the GCF

Factor out \(2x\) from \(10x^5 - 2x\): \(2x(5x^4 - 1)\)

Step3: Check if further factoring is possible

The expression \(5x^4 - 1\) is a difference of squares: \(5x^4 - 1 = (\sqrt{5}x^2)^2 - 1^2\). Using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), we get \((\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\). But usually, we factor over the integers. So the complete factorization over integers is \(2x(5x^4 - 1)\). If we consider factoring over reals, we can further factor \(5x^4 - 1\) as \((\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\) and \(\sqrt{5}x^2 - 1\) can be factored as \((x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\), but for basic factoring (over integers), the complete factorization is \(2x(5x^4 - 1)\). Wait, actually, \(5x^4-1\) can also be factored as a difference of squares in terms of \(x^2\) with rational coefficients? No, 5 is not a perfect square. Wait, maybe I made a mistake. Wait, the original expression: \(10x^5 - 2x = 2x(5x^4 - 1)\). Now, \(5x^4 - 1\) can be written as \((\sqrt{5}x^2)^2 - 1^2\), so factoring that gives \((\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\). And \(\sqrt{5}x^2 - 1\) can be factored as \((x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\) if we go to fourth roots, but typically, for high school level factoring (over integers), the complete factorization is \(2x(5x^4 - 1)\). Wait, no, maybe I missed that \(5x^4 - 1\) can be factored as a difference of squares with \(a = x^2\sqrt{5}\) and \(b = 1\), but if we are factoring over the real numbers, then yes. But the problem says "factor completely", so we need to factor until we can't factor anymore. So first, factor out the GCF: \(2x(5x^4 - 1)\). Then, \(5x^4 - 1\) is a difference of squares: \(a^2 - b^2=(a + b)(a - b)\) where \(a = \sqrt{5}x^2\) and \(b = 1\). So \(5x^4 - 1 = (\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\). Then, \(\sqrt{5}x^2 - 1\) is also a difference of squares: \(a^2 - b^2=(a + b)(a - b)\) where \(a = x\sqrt[4]{5}\) and \(b = 1\). So \(\sqrt{5}x^2 - 1=(x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\). So putting it all together: \(10x^5 - 2x = 2x(\sqrt{5}x^2 + 1)(x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\). But that seems complicated. Wait, maybe the problem expects factoring over integers, so the GCF is \(2x\), and then \(5x^4 - 1\) can't be factored over integers, so the complete factorization is \(2x(5x^4 - 1)\). Wait, no, let's check again. The original expression: \(10x^5 - 2x\). Factor out \(2x\): \(2x(5x^4 - 1)\). Now, \(5x^4 - 1\) is a difference of squares: \( (x^2\sqrt{5})^2 - 1^2\), so factoring that gives \((x^2\sqrt{5} + 1)(x^2\sqrt{5} - 1)\). Then, \(x^2\sqrt{5} - 1\) is also a difference of squares: \( (x\sqrt[4]{5})^2 - 1^2\), so factoring that gives \((x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\). So the complete factorization over the real numbers is \(2x(\sqrt{5}x^2 + 1)(x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\). But maybe the problem is at a level where we just factor out the GCF and then recognize \(5x^4 - 1\) as a difference of squares with irrational coefficients, but perhaps I made a mistake in the initial step. Wait, no, the GCF of 10 and 2 is 2, and the GCF of \(x^5\) and \(x\) is \(x\), so factoring out \(2x\) gives \(2x(5x^4 - 1)\). Then, \(5x^4 - 1\) can be factored as \((\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\) using the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\) where \(a = \sqrt{5}x^2\) and \(b = 1\). So that's the complete factorization.

Answer:

\(2x(5x^4 - 1)\) (or \(2x(\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\) or further factored as \(2x(\sqrt{5}x^2 + 1)(x\sqrt[4]{5} + 1)(x\sqrt[4]{5} - 1)\) depending on the domain of factoring, but the most common complete factorization (over integers for the GCF and then over reals for the quadratic in \(x^2\)) is \(2x(5x^4 - 1)\) or \(2x(\sqrt{5}x^2 + 1)(\sqrt{5}x^2 - 1)\))