Question

multiple$A=\frac{{3x}^4}{4}-\frac{x^4}{2}+5x-7;B=\frac{{2x}^3}{3}+\frac{{3x}^3}{2}-\frac{x}{6};C=\frac{{2x}^4}{3}-\frac{{3x}^4}{4}+5$ find
$BA$

Answer 1

Given that,

$A=\frac{3x^4}{4}-\frac{x^4}{2}+5x-7=\frac{x^4}{4}+5x-7$

$B=\frac{2x^3}{3}+\frac{3x^3}{2}-\frac{x}{6}=\frac{13x^3}{6}-\frac{x}{6}$

$C=\frac{2x^4}{3}-\frac{3x^4}{4}+5=-\frac{x^4}{12}+5$

$BA=(\frac{13x^3}{6}-\frac{x}{6})(\frac{x^4}{4}+5x-7)$

$=\frac{13x^7}{24}+\frac{65x^4}{6}-\frac{91x^3}{6}-\frac{x^5}{24}-\frac{5x^2}{6}+\frac{7x}{6}$

$=\frac{13x^7}{24}-\frac{x^5}{24}+\frac{65x^4}{6}-\frac{91x^3}{6}-\frac{5x^2}{6}+\frac{7x}{6}$