// Numbas version: exam_results_page_options {"name": "Factorising a non-monic quadratic", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Factorising a non-monic quadratic", "tags": ["binomial", "factorisation", "Factorisation", "factorise", "non-monic", "quadratic", "quadratics"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "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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$\\simplify{{c[1]}x^2+{d[1]+b[1]*c[1]}x+{b[1]*d[1]}}$ = [[0]].

\n

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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
  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\n
\n

As always, you can check your factorisation by expanding.

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Please factorise

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Please factorise

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

\n

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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
  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\n
\n

As always, you can check your factorisation by expanding.

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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