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A worked example using the quadratic expression from Part a:  $\\simplify{{f1}x^2+{f2}x+{f3}}$

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When the coefficient of $x^2$ is not $1$ in a quadratic expression such as this one it is a little more complicated.

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We must produce a pair of brackets  $(ax+c)(bx+d)$ which multiply out to give $\\simplify{{f1}x^2+{f2}x+{f3}}$

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Therefore we must find:

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the two factors, a and b,  of the coefficent of the $x^2$ term in the quadratic,  ($\\var{f1}$ in this example ) AND

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the two factors, c and d,  of the constant term in the quadratic,  ($\\var{f3}$ in this example),

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such that $(ax+c)(bx+d)$ = $\\simplify{{f1}x^2+{f2}x+{f3}}$

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Here we use $\\var{a}$ and $\\var{b}$ as the factors of $\\var{f1}$  and $\\var{c}$ and $\\var{d}$ as the factors of $\\var{f3}$

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and we must now combine them correctly such that $(ax+c)(bx+d)$$\\simplify{{f1}x^2+{f2}x+{f3}}$

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The correct combination is:  $\\simplify{({a}x+{c})({b}x+{d})}$

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These brackets, when multiplied out, will give the original quadratic expression $\\simplify{{f1}x^2+{f2}x+{f3}}$

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Try it!

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$\\simplify{{f1}x^2+{f2}*x+{f3}}$

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$=$[[0]]

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The answer must be factorised.

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The answer must be factorised.

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$\\simplify{{f12}x^2+{f22}*x+{f32}}$

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$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f13}x^2+{f23}*x+{f33}}$

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$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f14}x^2+{f24}*x+{f34}}$

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$=$[[0]] 

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The answer must be factorised.

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Express each of the following quadratic expressions as the product of two linear factors of the form $(ax+c)(bx+d)$.

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Factorising further basic quadratics into linear factors.

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