<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%5E2%3D16" id="TexFormula1" title="\sqrt{x} ^2=16" alt="\sqrt{x} ^2=16" align

$\bf x^4+xy^3+x^3+y^3\impliedby \textit{now let's do some grouping}
\\\\\\
(x^4+xy^3)~~+~~(x^3+y^3)\implies x(x^3+y^3)~~+~~(x^3+y^3)
\\\\\\
\stackrel{\stackrel{common}{factor}}{(x^3+y^3)}(x+1)\\\\
-------------------------------\\\\
\textit{now recall that }\qquad \qquad \textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\
-------------------------------\\\\
(x+y)(x^2-xy+y^2)(x+1)$

5 0

3 years ago

Can any negative integer be greater than a positive integer explain

Keith_Richards [23]

No, it is impossible. Intuitively, a negative number sits at the left of 0 on the number line, and a positive number sits at the right of 0 on the number line. And a number x is greater than another number y if x sits at the right of y on the number line. So, every positive number is greater than any negative number.

Also, by definition, a positive number is greater than 0, and a negative number is smaller than zero. So, if x is positive and y is negative, you have

$y < 0 < x$

and since the relation of order "<" is transitive, this implies

$y < x$

5 0

3 years ago