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Which of the following are factors of $\\simplify{{prime^primepower}x^{xpower}y^{ypower}(x+{xconstant})^{xconstantpower} (z+{failconstant})^{failsafepower}}$?

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Suppose that $\\var{factor1}${factor2} is a factor of an expression. What can be said of $-\\var{factor1}${factor2}?

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It is also a factor.

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It is not necessarily a factor.

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It is definitely not a factor.

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$\\simplify{{prime^primepower}x^{xpower}y^{ypower}(x+{xconstant})^{xconstantpower} (z+{failconstant})^{failsafepower}}$

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The expression $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is a sum and can be factorised (written as a product) by finding the largest common factor:

$\\var{pmult*pxcoeff}x+\\var{pconstant*pmult} = $ [[0]] $\\large($ [[1]] $\\large)$

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We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The (largest) common factor of $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ is $\\var{pmult}$. Once we remove that factor from each term in $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}$ we are left with $\\var{pxcoeff}x+\\var{pconstant}$.

That means $\\var{pmult*pxcoeff}x+\\var{pconstant*pmult}= \\var{pmult}(\\var{pxcoeff}x+\\var{pconstant})$.

Factorise $\\simplify{{bp1}a+{bp2}}$

[[0]] $\\large($ [[1]] $\\large)$

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We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The (largest) common factor of $\\simplify{{bp1}a+{bp2}}$ is $\\var{cf}$. Once we remove that factor from each term in $\\simplify{{bp1}a+{bp2}}$ we are left with $\\var{bx}a+\\var{bc}$.

That means $\\simplify{{bp1}a+{bp2}}$ is $\\var{cf} = \\var{cf}(\\var{bx}a+\\var{bc})$.

Factorise $\\simplify{{ct1}x+{ct2}y+{ct3}}$

[[0]] $\\large($ [[1]] $\\large)$

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We put the common factor out the front of a set of brackets and put the 'left-overs' inside.

The (largest) common factor of $\\simplify{{ct1}x+{ct2}y+{ct3}}$ is $\\var{cmult}$. Once we remove that factor from each term in $\\simplify{{ct1}x+{ct2}y+{ct3}}$ we are left with $\\simplify{{cx}x+{cy}y+{cc}}$.

That means $\\simplify{{ct1}x+{ct2}y+{ct3}} = \\simplify{{cmult}({ct1}x+{ct2}y+{ct3})}$.

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