Simon's copy of Factorising Quadratic Equations with $x^2$ Coefficients Greater than 1

Question 1

Factorise the equation

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

$(\var{a}x+\phantom{.}$Expected answer:$) ($Expected answer:$x+\phantom{.}$Expected answer:$)\; = 0$

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Write down the roots of the equation above.

Input your answer as $x_1$ and $x_2$, where $x_1<x_2$.

$x_1=$ Write your answer as a fraction.Expected answer:

$x_2=$ Write your answer as a fraction.Expected answer:

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This question is more difficult because we are factorising a quadratic of the form $ax^2+bx+c$ where $a \neq 1$.

To factorise an expression like $ax^2+bx+c$ we can start by finding two numbers which multiply to give $a\times c$ and add to give $b$.

In our example, to factorise $\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}$ we start by finding two numbers which:

Multiply to give $\var{a*b*c*d}$ $(a \times c)$ and add to give $\simplify{{a}*{d}+{b}*{c}}$ $(b)$

Note that $(\var{a*d}) \times (\var{b*c}) = \var{a*b*c*d}$

and

($\var{a*d})+(\var{b*c}) = \var{a*d+b*c}$

So the two numbers required are $\var{a*d}$ and $\var{b*c}$.

We can then return to our original quadratic and split the $x$ coefficient into two pieces, $\var{a*d}x$ and $\var{b*c}x$, as follows:

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

Now we can factorise the first two terms and the last two terms separately as:

$\simplify[!cancelterms]{{a*c}x^2+{a*d}x+{b*c}x+{b*d}} = \simplify[!cancelterms]{{a}x({c}x+{d})+{b}({c}x+{d})}$

which is the same as

$\simplify[!cancelterms]{({a}x+{b})({c}x+{d})}$ which is thus our final factorisation.

$\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\simplify{{a}x+{b}} = 0$ or $\simplify{{c}x+ {d}} = 0$.

So the roots of the equation are $\var[fractionnumbers]{-b/a}$ and $\var[fractionnumbers]{-d/c}$.