Question

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

Find $-2A+6B+3C$

Answer 1

Given that,

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

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

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

$-2A+6B+3C=-2(\frac{x^4}{4}+5x-7)+6(\frac{13x^3}{6}-\frac{x}{6})+3(-\frac{x^4}{12}+5)$

$=-\frac{x^4}{2}-10x+14+13x^3-x-\frac{x^4}{4}+15$

$=-\frac{3x^4}{4}+13x^3-11x+29$