3 questions on factorising quadratics. The second question also asks for the roots of the quadratic. The third question involves factorising quartic polynomials but which are quadratics in $x^2$.
", "licence": "Creative Commons Attribution 4.0 International"}, "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "shufflequestions": false, "questions": [], "percentpass": 0.0, "allQuestions": true, "pickQuestions": 0, "type": "exam", "feedback": {"showtotalmark": true, "advicethreshold": 0.0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": false, "showexpectedanswerswhen": "never", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Dividing polynomials 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["algebra", "algebraic fractions", "algebraic manipulation", "cancelling common terms in algebraic fractions", "factorising polynomials", "polynomials", "rational polynomials", "simplifying algebraic fractions"], "advice": "Note that both the numerator and denominator factorize and that they then have a common factor which can then be cancelled.
", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n \n \n$A=\\;\\;$[[0]]
\n \n \n \nInput all numbers as integers or fractions and not as decimals.
\n \n \n ", "gaps": [{"notallowed": {"message": "Reduce to its lowest form.
", "showstrings": false, "strings": ["^4", "^(4)"], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "((({m} * (x ^ 2)) + {(( - p) * q)}) / ((x ^ 2) + {n}))", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "statement": "\nSimplify the following expression:
\n\\[A=\\simplify[std]{({m} * x ^ 4 + {m -(p * q)} * x ^ 2 + {( -p) * q}) / (x ^ 4 + {1 + n} * x ^ 2 + {n})}\\]
\nYour answer should be as a single algebraic fraction in its lowest form.
\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"sq": {"definition": "random(1,-1)", "name": "sq"}, "sp": {"definition": "random(1,-1)", "name": "sp"}, "m": {"definition": "sm*random(1..5)", "name": "m"}, "n": {"definition": "sn*random(1..5)", "name": "n"}, "q": {"definition": "sq*random(1..5)", "name": "q"}, "p": {"definition": "sp*random(1..5)", "name": "p"}, "sn": {"definition": "random(1,-1)", "name": "sn"}, "sm": {"definition": "random(1,-1)", "name": "sm"}}, "metadata": {"notes": "\n \t\t27/06/2012:
\n \t\tMinor changes in display and statements/prompts.
\n \t\tAdded tags.
\n \t\tCould have more detailed advice.
\n \t\t18/07/2012:
\n \t\tAdded description.
\n \t\t23/07/2012:
\n \t\tAdded tags.
\n \t\tThe Advice section is very brief - Should there be more help provided here?
\n \t\tQuestion appears to be working correctly.
\n \t\t", "description": "Algebraic manipulation/simplification.
\nSimplify $\\displaystyle \\frac{ax^4+bx^2+c}{a_1x^4+b_1x^2+c_1}$ by cancelling a a common degree 2 factor.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Factorise a quadratic", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Steps", "factorisation", "factorise a quadratic", "factorization", "factorize a quadratic", "linear factors", "quadratics", "steps"], "advice": "\nDirect 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"}], "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.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Solving equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Steps", "algebra", "factorisation", "find roots of a quadratic equation", "quadratic formula", "quadratics", "roots of a quadratic equation", "solving a quadratic equation", "solving equations", "steps"], "advice": "\nDirect 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})}\\]
\nHence we find the roots:
\\[\\begin{eqnarray} x&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ x&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]
Other Methods.
\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 $\\displaystyle -\\frac{b}{2a}$
\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 $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$
\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}\\]
\nCompleting 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}\\]
Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]
You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.
\nEnter the least root first. If the roots are equal, enter the root in both input boxes.
\nEnter the roots as fractions or integers, not as decimals.
\n ", "gaps": [{"notallowed": {"message": "Input numbers as fractions or integers not as a decimals.
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.0001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{n1-n4}/{2*a*b}", "type": "jme"}, {"notallowed": {"message": "Input numbers as fractions or integers not as a decimals.
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.0001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{n1+n4}/{2*a*b}", "type": "jme"}], "steps": [{"prompt": "\nFinding the roots by factorisation.
\nFinding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immendiately.
\nIf you cannot find a factorisation then there are several other methods you can use.
\nUsing the formula for the roots.
\nYou can find the roots by using the 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 value is $\\displaystyle \\frac{-b}{2a}$
\n3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.
\n\n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "statement": "
Find the roots of the following quadratic equation.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..5)", "name": "a"}, "c": {"definition": "c1*s3", "name": "c"}, "b": {"definition": "random(1..4)", "name": "b"}, "d": {"definition": "if((a*d1)^2=(b*c)^2, max(d1+1,random(1..5))*s3,d1*s3)", "name": "d"}, "f": {"definition": "a*b", "name": "f"}, "s3": {"definition": "random(1,-1)", "name": "s3"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "n5": {"definition": "a*b", "name": "n5"}, "disc": {"definition": "(b*c+a*d)^2-4*a*b*c*d", "name": "disc"}, "rep": {"definition": "switch(disc=0,'repeated', ' ')", "name": "rep"}, "rdis": {"definition": "switch(disc=0,'The discriminant is '+ 0+' and so we get two repeated roots in this case.',disc<0, 'There are no real roots.','The roots exist and are distinct. ')", "name": "rdis"}, "n1": {"definition": "b*c+a*d", "name": "n1"}, "c1": {"definition": "switch(f=1, random(1..6),f=2,random(1,3,5,7,9),f=3,random(1,2,5,7,8),f=4,random(1,3,5,7,9),f=6, random(1,5,7,8),f=9,random(1,2,4,7,8),f=8,random(1,3,5,7,9),f=12,random(1,5,7),random(1,3,5,7))", "name": "c1"}, "n3": {"definition": "2*a*b", "name": "n3"}, "n4": {"definition": "abs(n2)", "name": "n4"}, "n2": {"definition": "b*c-a*d", "name": "n2"}, "d1": {"definition": "switch(f=1, random(1..6),f=2,random(1,3,5,7,9),f=3,random(1,2,5,7,8),f=4,random(1,3,5,7,9),f=6, random(1,5,7,8),f=9,random(1,2,4,7,8),f=8,random(1,3,5,7,9),f=12,random(1,5,7),random(1,3,5,7))", "name": "d1"}}, "metadata": {"notes": "\n \t\t5/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tImproved display in various content areas.
\n \t\t", "description": "Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "extensions": [], "custom_part_types": [], "resources": []}