Answer: No, [tex]Product\ of\ \frac{1}{2}x-\frac{1}{4}\ and\ 5x^2-2x+6\text{ is not equal to Product of }\frac{1}{4}x-\frac{1}{2}\ and\ 5x^2-2x+6[/tex]
Step-by-step explanation:
Since we have given that
1) Product of [tex]\frac{1}{2}x-\frac{1}{4}\ and\ 5x^2-2x+6[/tex]
So, the product becomes,
[tex](\frac{1}{2}x-\frac{1}{4})(5x^2-2x+6)\\\\=\frac{x}{2}(5x^2-2x+6)-\frac{1}{4}(5x^2-2x+6)\\\\=\frac{5x^3}{2}-x^2+3x-\frac{5x^2}{4}+\frac{1x}{2}-\frac{3}{2}\\\\=\frac{5x^3}{2}-\frac{5x^2}{4}-x^2+3x+\frac{x}{2}-\frac{3}{2}\\\\=\frac{5x^3}{2}-\frac{9x^2}{4}+\frac{7x}{2}-\frac{3}{2}[/tex]
2)Product of [tex]\frac{1}{4}x-\frac{1}{2}\ and\ 5x^2-2x+6[/tex]
is given by
[tex](\frac{1}{4}x-\frac{1}{2})(5x^2-2x+6)\\\\=\frac{x}{4}(5x^2-2x+6)-\frac{1}{2}(5x^2-2x+6)\\\\=\frac{5x^3}{4}-\frac{1}{2}x^2+\frac{3x}{2}-\frac{5x^2}{2}+x-3\\\\=\frac{5x^3}{4}-\frac{6x^2}{2}+\frac{5x}{2}-3\\\\=\frac{5x^3}{4}-3x^2+\frac{5x}{2}-3[/tex]
No, Product of [tex]\frac{1}{2}x-\frac{1}{4}\ and\ 5x^2-2x+6[/tex] is not equal to Product of [tex]\frac{1}{4}x-\frac{1}{2}\ and\ 5x^2-2x+6[/tex]
