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Factorise the following into squares, that is, $(ax+b)^2$ or if there is a common factor $k$, then $k(cx+d)^2$.

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$\\simplify{{aa}x^2+{mid}x+{bb}}$ = [[0]].

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Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

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In fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since

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That is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.

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Each bracket has a common factor of $\\var{g}$, so we call move both of them to the front, to get

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$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.

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You can always 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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$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ = [[0]].

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Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.

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In fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since

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That is, 

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$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.

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You can always 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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