// Numbas version: exam_results_page_options {"name": "Terry's copy of Solving a non-monic quadratic by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "", "statement": "

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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Please factorise

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Please factorise

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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{{c[1]}x^2+{d[1]+b[1]*c[1]}x+{b[1]*d[1]}}$$=$0
$\\Longrightarrow$ [[0]]$=$0
$\\Longrightarrow$$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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There are a few different ways to do the working for these questions, here is one method that uses factorisation by grouping.

\n


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

\n
    \n
  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
  2. multiply the constant term and the coefficient of $x^2$ to get $\\var{c[1]/{g_one}*{b[1]*d[1]/g_one}}$
    3. \n
  3. 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 $\\var{d[1]/g_one}$
    4. \n
  4. Use these numbers to decompose the coefficient of the $x$ term, $\\simplify{{g_one}({c[1]/g_one}x^2+{b[1]*c[1]/g_one}x+{d[1]/g_one}x+{b[1]*d[1]/g_one})}$
    5. \n
  5. Use factorisation by grouping to factorise the quadratic, that is:
    6. \n\n
\n

Now, since $\\simplify{{g_one}(x+{b[1]})({c[1]/g_one}x + {d[1]/g_one}  )}=0$, by the null factor law, either 

\n

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

\n

Solving these equations results in 

\n

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

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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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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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

\n
    \n
  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
  2. multiply the constant term and the coefficient of $x^2$ to get $\\var{a[0]*c[0]*b[0]*d[0]/(gab_zero*gcd_zero)^2}$
    3. \n
  3. 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 $\\var{(b[0]*c[0])/(gab_zero*gcd_zero)}$
    4. \n
  4. Use these numbers to decompose the coefficient of the $x$ term, $\\simplify{{gab_zero*gcd_zero}({a[0]*c[0]/(gab_zero*gcd_zero)}x^2+{(a[0]*d[0])/(gab_zero*gcd_zero)}x+{(b[0]*c[0])/(gab_zero*gcd_zero)}x+{b[0]*d[0]/(gab_zero*gcd_zero)})}$
    5. \n
  5. Use factorisation by grouping to factorise the quadratic, that is:
    6. \n\n
\n

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 

\n

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

\n

Solving these equations results in 

\n

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

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I could use !noLeadingMinus in simplify to avoid it rearranging

", "description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "tags": ["binomial", "factorisation", "Factorisation", "factorise", "non-monic", "quadratic", "quadratics", "solving"], "rulesets": {}, "ungrouped_variables": ["a", "b", "c", "d", "g_one", "gab_zero", "gcd_zero"], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Terry Young", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3130/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Terry Young", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3130/"}]}