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Factorise three quadratic equations of the form $x^2+bx+c$.

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The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

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\$x^2+bx+c=0\$

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can be factorised to create an equation of the form

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\$(x+m)(x+n)=0\\text{.}\$

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When we expand a factorised quadratic expression we obtain

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\$(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\$

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To factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

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#### a)

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\$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\$

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We need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.

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\\\begin{align} \\var{v1} \\times \\var{v2}&=\\var{v1*v2}\\\\ \\var{v1}+\\var{v2}&=\\var{v1+v2}\\\\ \\end{align} \

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So the factorised form of the equation is

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\$\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\$

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#### b)

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We can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.

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\\\begin{align} \\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\ \\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\ \\end{align} \

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So the factorised form of the equation is

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\$\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\$

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#### c)

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\$\\simplify{x^2+{v5*v6}=0}\$

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we need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.

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\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}

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So the factorised form of the equation is

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\$\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\$

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

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[[0]] $=0$

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