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Unit 1: Sequences and Functions
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Factorise the following cubic expression into a product of linear factors, where one of the factors is $\\simplify{{b1}x+{c1}}$:
\n\\[\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}} \\]
", "advice": "If the cubic expression
\n\\[\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}} \\]
\ncan be factorised into a product of linear factors, and we know that one factor is $\\simplify{{b1}x+{c1}}$, it is easiest to first find the remaining quadratic factor. We can then factorise this quadratic into two linear factors.
\nWe can start by taking a generalised quadratic in the form $ax^2+bx+c$ as the quadratic factor:
\n\\[ \\begin{split}\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}} &\\,= (\\simplify{{b1}x+{c1}})(ax^2+bx+c) \\\\ &\\,= \\simplify{{b1}x(a*x^2+b*x+c) + {c1}(a*x^2+b*x+c)}. \\end{split} \\]
\nIf we now expand the brackets and collect similar terms, we can compare coefficients to find the values of $a$, $b$, and $c$:
\n\\[ \\begin{split}\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}} &\\,= \\simplify{{b1}x(a*x^2+b*x+c) + {c1}(a*x^2+b*x+c)} \\\\ &\\,=\\simplify[all,!collectNumbers, !cancelTerms]{{b1}a x^3 + {b1}b x^2 +{b1} c x + {c2}a x^2 + {c2} b x + {c2} c} \\\\ &\\,=\\simplify{{b1}a x^3 + ({b1}b+{c1}a)x^2+({b1}c+{c1}b)x +{c1}c }.\\end{split} \\]
\nBy comparing the coefficients of each side of the equation we can create a set of simultaneous equations to find $a$, $b$, and $c$:
\n\\[ \\begin{split} & \\simplify{{b1*b2*b3}} &\\,= \\simplify{{b1}a} \\\\ &\\simplify{{c1*b2*b3+c2*b1*b3+c3*b1*b2}} &\\,= \\simplify{{b1}b+{c1}a} \\\\ &\\simplify{{b1*c2*c3+b2*c1*c3+b3*c1*c2}} &\\,=\\simplify{{b1}c+{c1}b} \\\\ &\\simplify{{c1*c2*c3}} &\\,=\\simplify{{c1}c}. \\end{split} \\]
\nHence, \\[ a=\\var{b2*b3},\\quad b=\\var{b1},\\quad c=\\var{c1}.\\]
\nTherefore, \\[\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}} = (\\simplify{{b1}x+{c1}})(\\simplify{{b2*b3}x^2+{c2*b3+b2*c3}x+{c2*c3}}). \\]
\nFinally, we need to factorise the quadratic $\\simplify{{b2*b3}x^2+{c2*b3+b2*c3}x+{c2*c3}}$:
\n\\[ \\simplify{{b2*b3}x^2+{c2*b3+b2*c3}x+{c2*c3}}=(\\simplify{{b2}x+{c2}})(\\simplify{{b3}x+{c3}}).\\]
\nThus, the cubic expression as a product of three linear factors is
\n\\[\\simplify{{b1*b2*b3}x^3+{c1*b2*b3+c2*b1*b3+c3*b1*b2}x^2+{b1*c2*c3+b2*c1*c3+b3*c1*c2}x+{c1*c2*c3}}=(\\simplify{{b1}x+{c1}})(\\simplify{{b2}x+{c2}})(\\simplify{{b3}x+{c3}}). \\]
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