## Algebra 2 (1st Edition)

Substituting each of the numbers into the equation. $f(-\frac{3}{2})\\=40(-\frac{3}{2})^5-42(-\frac{3}{2})^4-107(-\frac{3}{2})^3+107(-\frac{3}{2})^2+33(-\frac{3}{2})-36\\=40(-\frac{243}{2})-42(\frac{81}{16})-107(-\frac{27}{8})+107(\frac{9}{4})+33(-\frac{3}{2})-36\\=0$ So, $-\frac{3}{2}$ is a zero of the function. $f(-\frac{3}{8})\\=40(-\frac{3}{8})^5-42(-\frac{3}{8})^4-107(-\frac{3}{8})^3+107(-\frac{3}{8})^2+33(-\frac{3}{8})-36\\=40(-\frac{243}{32768})-42(\frac{81}{4096})-107(-\frac{27}{512})+107(\frac{9}{64})+33(-\frac{3}{8})-36\\\approx-28.8$ So, $-\frac{3}{8}$ is not a zero of the function. $f(\frac{3}{4})\\=40(\frac{3}{4})^5-42(\frac{3}{4})^4-107(\frac{3}{4})^3+107(\frac{3}{4})^2+33(\frac{3}{4})-36\\=40(\frac{243}{1024})-42(\frac{81}{256})-107(\frac{27}{64})+107(\frac{9}{16})+33(\frac{3}{4})-36\\=0$ So, $\frac{3}{4}$ is a zero of the function. $f(\frac{4}{5})\\=40(\frac{4}{5})^5-42(\frac{4}{5})^4-107(\frac{4}{5})^3+107(\frac{4}{5})^2+33(\frac{4}{5})-36\\=40(\frac{1024}{3125})-42(\frac{256}{625})-107(\frac{64}{125})+107(\frac{16}{25})+33(\frac{4}{5})-36\\=0$ So, $\frac{4}{5}$ is a zero of the function.