## Precalculus (6th Edition) Blitzer

We know that by using Pascal’s Triangle, we can find the coefficient of each term without expansion of the given function. Therefore, every term having the odd power of $\left( -1 \right)$ will be negative and the terms having even powers of $\left( -1 \right)$ will be positive. Therefore, the first, third, fifth and seventh terms will be positive. And the second, fourth and sixth terms will be negative. Rewrite the function ${{\left( x-1 \right)}^{6}}$ as ${{\left( x+\left( -1 \right) \right)}^{6}}$. Then, by applying Pascal’s Triangle: Positive terms: ${{x}^{6}},15{{x}^{4}}{{\left( -1 \right)}^{2}},15{{x}^{2}}{{\left( -1 \right)}^{4}},{{\left( -1 \right)}^{6}}$ After solving the terms: ${{x}^{6}},15{{x}^{4}},15{{x}^{2}},1$ Negative terms: $6{{x}^{5}}{{\left( -1 \right)}^{1}},20{{x}^{3}}{{\left( -1 \right)}^{3}},6{{x}^{1}}{{\left( -1 \right)}^{5}}$ Then, after solving the terms: $-6{{x}^{5}},-20{{x}^{3}},-6{{x}^{1}}$ Thus, the statement makes sense and the positive and negative terms are ${{x}^{6}},15{{x}^{4}},15{{x}^{2}},1,-6{{x}^{5}},-20{{x}^{3}},-6{{x}^{1}}$.