Factorise three quadratic equations of the form $x^2+bx+c$.
\nThe first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Factorise the following quadratic equations.
\n", "advice": "Quadratic equations of the form
\n\\[x^2+bx+c=0\\]
\ncan be factorised to create an equation of the form
\n\\[(x+m)(x+n)=0\\text{.}\\]
\nWhen we expand a factorised quadratic expression we obtain
\n\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]
\nTo 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 $c$.
\n\\[\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\\]
\nWe need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.
\n\\[\\begin{align}
\\var{v1} \\times \\var{v2}&=\\var{v1*v2}\\\\
\\var{v1}+\\var{v2}&=\\var{v1+v2}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\\]
\n\nWe can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.
\n\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]
\nWhen factorising the quadratic expression
\n\\[\\simplify{x^2+{v5*v6}=0}\\]
\nwe need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.
\n\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}
So the factorised form of the equation is
\n\\[\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\\]
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\n[[0]] $=0$
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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$
\n[[0]] $=0$
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$\\simplify{x^2+{v5*v6}}=0$
\n[[0]] $=0$
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