// Numbas version: exam_results_page_options {"name": "Solving a non-monic quadratic by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "c", "d", "g_one", "gab_zero", "gcd_zero"], "name": "Solving a non-monic quadratic by factorising", "tags": ["binomial", "factorisation", "Factorisation", "factorise", "non-monic", "quadratic", "quadratics", "solving"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": 0, "prompt": "

Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 \$\\simplify{{c[1]}x^2+{d[1]+b[1]*c[1]}x+{b[1]*d[1]}}\$ \$=\$ 0 \$\\Longrightarrow\$ [[0]] \$=\$ 0 \$\\Longrightarrow\$ \$x\$ \$=\$ [[1]]
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Note: In the first gap, enter the quadratic in factored form.

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Note: In the second gap, if \$x=1,2\$, enter set(1,2).

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Please factorise

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Please factorise

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There are a few different ways to do the working for these questions, here is one method that uses factorisation by grouping.

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Given \$\\simplify{{c[1]}x^2+{d[1]+b[1]*c[1]}x+{b[1]*d[1]}}\$, we

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1. look for a common factor, in this case it is \$\\var{g_one}\$, and put it out the front: \$\\simplify{{g_one}({c[1]/g_one}x^2+{(d[1]+b[1]*c[1])/g_one}x+{b[1]*d[1]/g_one})}\$
2. \n
3. multiply the constant term and the coefficient of \$x^2\$ to get \$\\var{c[1]/{g_one}*{b[1]*d[1]/g_one}}\$
4. \n
5. find two numbers that multiply to give \$\\var{c[1]/g_one*b[1]*d[1]/g_one}\$ and add to give \$\\var{(d[1]+b[1]*c[1])/g_one}\$, in this case the numbers are \$\\var{b[1]*c[1]/g_one}\$ and \$\\var{d[1]/g_one}\$
6. \n
7. Use these numbers to decompose the coefficient of the \$x\$ term, \$\\simplify{{g_one}({c[1]/g_one}x^2+{b[1]*c[1]/g_one}x+{d[1]/g_one}x+{b[1]*d[1]/g_one})}\$
8. \n
9. Use factorisation by grouping to factorise the quadratic, that is:
10. \n
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• find the common factor for the first two terms, and then the second two terms, \$\\simplify{{g_one}({c[1]/g_one}x(x+{b[1]}) + {d[1]/g_one}(x+{b[1]})  )}\$
• \n
• realise \$\\simplify{x+{b[1]}}\$ is itself a common factor and take that out the front, \$\\simplify{{g_one}(x+{b[1]})({c[1]/g_one}x + {d[1]/g_one}  )}\$
• \n
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Now, since \$\\simplify{{g_one}(x+{b[1]})({c[1]/g_one}x + {d[1]/g_one}  )}=0\$, by the null factor law, either

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\$\\simplify{x+{b[1]}}=0\$, or \$\\simplify{{c[1]/g_one}x + {d[1]/g_one}  =0}\$.

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Solving these equations results in

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\$x=\\var{-b[1]}\$, or \$x=\\simplify{{-d[1]}/{c[1]}}\$.

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Solve the following quadratic by factorisation:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 \$\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+{b[0]*d[0]}}\$ \$=\$ 0 \$\\Longrightarrow\$ [[0]] \$=\$ 0 \$\\Longrightarrow\$ \$x\$ \$=\$ [[1]]
\n

\n

Note: In the first gap, enter the quadratic in factored form.

\n

Note: In the second gap, if \$x=1,2\$, enter set(1,2).

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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There are a few different ways to do the working for these questions, here is one method that uses factorisation by grouping.

\n

Given \$\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+{b[0]*d[0]}}\$, we

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1. look for a common factor, in this case it is \$\\var{gab_zero*gcd_zero}\$, and put it out the front:  \$\\simplify{{gab_zero*gcd_zero}({a[0]*c[0]/(gab_zero*gcd_zero)}x^2+{(a[0]*d[0]+b[0]*c[0])/(gab_zero*gcd_zero)}x+{b[0]*d[0]/(gab_zero*gcd_zero)})}\$
2. \n
3. multiply the constant term and the coefficient of \$x^2\$ to get \$\\var{a[0]*c[0]*b[0]*d[0]/(gab_zero*gcd_zero)^2}\$
4. \n
5. find two numbers that multiply to give \$\\var{a[0]*c[0]*b[0]*d[0]/(gab_zero*gcd_zero)^2}\$ and add to give \$\\var{(a[0]*d[0]+b[0]*c[0])/(gab_zero*gcd_zero)}\$, in this case the numbers are \$\\var{(a[0]*d[0])/(gab_zero*gcd_zero)}\$ and \$\\var{(b[0]*c[0])/(gab_zero*gcd_zero)}\$
6. \n
7. Use these numbers to decompose the coefficient of the \$x\$ term, \$\\simplify{{gab_zero*gcd_zero}({a[0]*c[0]/(gab_zero*gcd_zero)}x^2+{(a[0]*d[0])/(gab_zero*gcd_zero)}x+{(b[0]*c[0])/(gab_zero*gcd_zero)}x+{b[0]*d[0]/(gab_zero*gcd_zero)})}\$
8. \n
9. Use factorisation by grouping to factorise the quadratic, that is:
10. \n
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• find the common factor for the first two terms, and then the second two terms, \$\\simplify{{gab_zero*gcd_zero}({a[0]/(gab_zero)}x({c[0]/gcd_zero}x+{d[0]/gcd_zero}) + {b[0]/(gab_zero)}({c[0]/gcd_zero}x+{d[0]/gcd_zero})  )}\$
• \n
• realise \$\\simplify{{c[0]/gcd_zero}x+{d[0]/gcd_zero}}\$ is itself a common factor and take that out the front, \$\\simplify{{gab_zero*gcd_zero}({c[0]/gcd_zero}x+{d[0]/gcd_zero})({a[0]/(gab_zero)}x+{b[0]/(gab_zero)}  )}\$
• \n
\n
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Now, since \$\\simplify{{gab_zero*gcd_zero}({c[0]/gcd_zero}x+{d[0]/gcd_zero})({a[0]/(gab_zero)}x+{b[0]/(gab_zero)}  )}=0\$, by the null factor law, either

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\$\\simplify{{c[0]/gcd_zero}x+{d[0]/gcd_zero}}=0\$, or \$\\simplify{{a[0]/(gab_zero)}x+{b[0]/(gab_zero)}  =0}\$.

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Solving these equations results in

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\$x=\\simplify{{-d[0]}/{c[0]}}\$, or \$x=\\simplify{{-b[0]}/{a[0]}}\$.

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Factorise the following into linear factors. That is, write the quadratic as a product of terms that look like \$ax+b\$ where \$a\$ and \$b\$ are real numbers.

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I could use !noLeadingMinus in simplify to avoid it rearranging

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