Simplify the following algebraic expressions.
\nNote: Although the question may accept coefficients in their decimal forms, it would be more appropriate to keep them in their most simplified fraction forms.
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\nAs before, factorise the numerator and denominator. This time, however, you'll notice that the factors themselves are the same.
\n$\\big(\\frac{\\var{n2}n}{{\\var{n2}}n}\\big)\\big(\\frac{\\var{b1}n+\\var{b2}}{\\var{b3}n+\\var{b4}}\\big)$
\nThe factors cancel, leaving:
\n$\\big(\\frac{\\var{b1}n+\\var{b2}}{\\var{b3}n+\\var{b4}}\\big)$
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\nHere, the quadratic expression in the numerator needs to be factorised into the product of two binomials.
\n$\\frac{({\\simplify{x+{c1}}})({\\simplify{x+{c2}}})}{({\\simplify{x+{c1}}})}$
\nYou will notice that one of the binomials in the numerator is the same as the denominator, which means that they can be cancelled. This leaves the expression:
\n${\\simplify{x+{c2}}}$
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\nThis time there is a quadratic expression which needs to be factorised into the products of binomials in both the numerator and denominator.
\n$\\frac{({\\simplify{n+{e1}}})({\\simplify{n+{e2}}})}{({\\simplify{n+{e1}}})({\\simplify{n+{e3}}})}$
\nThe repeated binomials in the numerator and denominator cancel, leaving:
\n$\\frac{({\\simplify{n+{e2}}})}{({\\simplify{n+{e3}}})}$
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\nFor this question, start by factorising each fraction being multiplied.
\n$\\big(\\var{co1}\\big)\\big(\\frac{\\simplify{x+{f1}}}{\\simplify{x+{f2}}}\\big)\\times\\big(\\var{co2}\\big)\\big(\\frac{\\simplify{x+{f2}}}{\\simplify{x+{f1}}}\\big)$
\nDue to the commmutative nature of multiplication, the factors can be rearranged so that potential simplification becomes easier to spot.
\n$\\big(\\var{co1}\\times\\var{co2}\\big)\\Big(\\frac{(\\simplify{x+{f1}})(\\simplify{x+{f2}})}{(\\simplify{x+{f2}})(\\simplify{x+{f1}})}\\Big)$
\nThe binomial expressions in the fraction all cancel, leaving the answer as the product of the factorised coefficients:
\n$\\var{co1}\\times\\var{co2}=\\simplify{{co1}*{co2}}$
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\nWhen the divisor is a fraction, it's often useful to flip the numerator and denominator and change the operation to multiplication. This 'flipping' action is more commonly referred to as taking the reciprocal and it holds that dividing by a number or expression is the same as multiplying by its reciprocal.
\n$\\simplify{{co3}/(x^({p3}+1) + {g1}x^{p3})}\\times \\simplify{({g2}x+{g2}{g1})/{co4}}$
\nEach fraction should then be separately factorised.
\n$\\big(\\frac{\\var{co3}}{x^{\\var{p3}}}\\big)\\Big(\\frac{1}{\\simplify{x+{g1}}}\\Big)\\times\\big(\\frac{\\var{g2}}{\\var{co4}}\\big)\\Big(\\frac{\\simplify{x+{g1}}}{1}\\Big)$
\nRearrange the factors to help spot simplification possibilities.
\n$\\big(\\frac{\\var{co3}}{x^{\\var{p3}}}\\big)\\big(\\frac{\\var{g2}}{\\var{co4}}\\big)\\times\\Big(\\frac{1}{\\simplify{x+{g1}}}\\Big)\\Big(\\frac{\\simplify{x+{g1}}}{1}\\Big)$
\nWhen multiplied, the final two fractions simplify and the binomials cancel.
\n$\\big(\\frac{\\var{co3}}{x^{\\var{p3}}}\\big)\\big(\\frac{\\var{g2}}{\\var{co4}}\\big)\\times\\Big(\\frac{\\simplify{x+{g1}}}{\\simplify{x+{g1}}}\\Big)=\\big(\\frac{\\var{co3}}{x^{\\var{p3}}}\\big)\\big(\\frac{\\var{g2}}{\\var{co4}}\\big)\\times1$
\nThis leaves the answer as the simplification of the remaining fractions multiplied.
\n$\\big(\\frac{\\var{co3}}{x^{\\var{p3}}}\\big)\\big(\\frac{\\var{g2}}{\\var{co4}}\\big)=\\simplify{({co3}*{g2})/({co4}*x^{p3})}$
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", "type": "question", "contributors": [{"name": "Jane Courtney", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2154/"}]}]}], "contributors": [{"name": "Jane Courtney", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2154/"}]}