Direct Factorisation.
\nIf you can spot a direct factorisation then this is the quickest way to do this question.
\nFor this example we have the factorisation
\n\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]
\nFactorisation by finding the roots.
\nFor if $x=r$ and $x=s$ and are the roots then $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$.
\nThere are several methods of finding the roots – here are the main methods.
\nFinding the roots of a quadratic using the standard formula.
\nWe can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$
\nThe two roots are
\n\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$
1. $\\Delta \\gt 0$. The roots are real and distinct
\n2. $\\Delta=0$. The roots are real and equal. Their common value is $-\\frac{b}{2a}$
\n3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.
\nFor this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify{{-(b*c+a*d)}^2-4*{a*b}*{c*d}={disc}}$
\n{rdis}.
\nSo the {rep} roots are:
\n\\[\\begin{eqnarray} x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 + n4}/ {n3}}\\\\ x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{(\\var{n1} - \\var{n4}) }{\\var{n3}} &=& \\simplify{{n1 - n4}/ {n3}} \\end{eqnarray}\\]
So we see that:
\\[q(x)=\\simplify{{a*b}}\\left(\\simplify{x-{n1 + n4}/ {n3}}\\right)\\left(\\simplify{x-{n1 - n4}/ {n3}}\\right)=\\simplify{({b} * x + { -d}) * ({a} * x + { -c})}\\]
Completing the square.
\nFirst we complete the square for the quadratic expression $\\simplify{{a*b}x^2+{-n1}x+{c*d}}$
\\[\\begin{eqnarray} \\simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\\var{n5}\\left(\\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2+ \\simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2 -\\simplify{ {n2^2}/{4*(a*b)^2}}\\right) \\end{eqnarray} \\]
So to solve $\\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:
\\[\\begin{eqnarray} \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}& -\\simplify{ {n2^2}/{4*(a*b)^2}}=0\\Rightarrow\\\\ \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}&=\\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \\end{eqnarray}\\]
So we get the two {rep} solutions:
\\[\\begin{eqnarray} \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{{abs(n2)}/{2*a*b}} \\Rightarrow &x& = \\simplify{({abs(n2)+n1}/{2*a*b})}\\\\ \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{-({abs(n2)}/{2*a*b})} \\Rightarrow &x& = \\simplify{({n1-abs(n2)}/{2*a*b})} \\end{eqnarray}\\]
Finding these roots then gives the factorisation as before.
\\[q(x)=\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}\\]
$q(x)=\\;$ [[0]]
You can get more information on factorising a quadratic by clicking on Show steps. You will lose 1 mark if you do so.
\n ", "gaps": [{"notallowed": {"message": "
Factorise the expression into two factors.
", "showstrings": false, "strings": ["^", "x*x", "x x", "x(", "x (", ")x", ") x"], "partialcredit": 0.0}, "checkingaccuracy": 0.0001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 2.0, "answer": "((({a} * x) + {( - c)}) * (({b} * x) + {( - d)}))", "type": "jme", "musthave": {"message": "factorise the expression into two factors
", "showstrings": false, "strings": ["(", ")"], "partialcredit": 0.0}}], "steps": [{"prompt": "\nFactorisation by finding the roots
\nIf you cannot spot a direct factorisation of a quadratic $q(x)$ then finding the roots of the equation $q(x)=0$ can help you.
\nFor if $x=r$ and $x=s$ and are the roots then $q(x)=a(x-r)(x-s)$ for some constant $a$.
\nFinding the roots of a quadratic using the standard formula
We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$
The two roots are
\n\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$
1. $\\Delta \\gt 0$. The roots are real and distinct
\n2. $\\Delta=0$. The roots are real and equal. Their value is $\\frac{-b}{2a}$
\n3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.
\n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "extensions": [], "statement": "Factorise the following quadratic expression $q(x)$ into linear factors i.e. input $q(x)$ in the form
\\[(ax+b)(cx+d)\\] for suitable integers $a$, $b$, $c$ and $d$ .
5/08/2012:
\n \t\tAdded more tags.
\n \t\tAdded description.
\n \t\tAllowed the use of decimals.
\n \t\tImproved display of Advice.
\n \t\t", "description": "Factorise $\\displaystyle{ax ^ 2 + bx + c}$ into linear factors.
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