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Questions asking for the factoring of simple case (a=1)

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Then using factors to find roots.

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A quadratic equation in \"Standard\" form looks like this:

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\\( \\color{red}{a} x^2 + \\color{red}{b} x + \\color{red}{c} = 0 \\)

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Where \\( a, b \\) and \\( c \\)  are numerical co-efficients. \"\\( a \\)\" cannot be \\( 0 \\)   but  \"\\( b \\)\" and \"\\( c \\)\" can be any value (including \\( 0 \\)). The equation MUST be in this form before applying ANY method for solving.

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Some quadratic equations will factorise giving a simple way to \"solve\" them and find their roots. In the \"simple case\"  (when \\( a=1 \\)) this can be acheived by finding factors whose product (\\( \\times \\)) is \\( c \\) and sum (\\( + \\)) is \\( b \\).

", "advice": "

We are asked to factor different quadratic equations.

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The first step is to check that the equations are in \"standard form\", in these questions, they are. Also, these equations are all \"simple case\" quadratics (the co-efficient of \\( x^2 \\) is one), i.e. \\( a=1 \\).

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We know that, to get the \\( x^2 \\) term we need a single  \\( x \\) in each bracket:

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\\(  ( x + ??)( x + ?? )  \\)

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Then, as stated earlier, you just need to find a pair of numbers that:

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Once the equation is factored, we have  \\(  ( x + ??)( x + ?? ) =0 \\)

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The only way that this is possible is if one (or both) of the brackets equals zero.

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a)     \\( \\simplify{  x^2 + {b1}x + {c1} =0  } \\)

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We know that, to get the \\( x^2 \\) term we need a single  \\( x \\) in each bracket:

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\\(  ( x + ??)( x + ?? ) =0  \\)

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Then, we need to find two numbers that:

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A little thought leads us to  \\( \\var{f1} \\)  and  \\( \\var{f2} \\). these can be inserted into the brackets to give:

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\\(  \\simplify{ ( x + {f1})( x + {f2} ) }=0  \\)

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Now

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If  \\(  \\simplify{ ( x + {f1}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{f1} } \\)

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If  \\(  \\simplify{ ( x + {f2}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{f2} } \\)

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b)     \\( \\simplify{  x^2 + {b2}x + {c2} =0  } \\)

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We know that, to get the \\( x^2 \\) term we need a single  \\( x \\) in each bracket:

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\\(  ( x + ??)( x + ?? ) =0  \\)

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Then, we need to find two numbers that:

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A little thought leads us to  \\( \\var{g1} \\)  and  \\( \\var{g2} \\). these can be inserted into the brackets to give:

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\\( \\simplify{  ( x + {g1})( x + {g2} ) =0 } \\)

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Now

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If  \\(  \\simplify{ ( x + {g1}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{g1} } \\)

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If  \\(  \\simplify{ ( x + {g2}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{g2} } \\)

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c)     \\( \\simplify{  x^2 + {b3}x + {c3} =0  } \\)

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We know that, to get the \\( x^2 \\) term we need a single  \\( x \\) in each bracket:

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\\(  ( x + ??)( x + ?? ) =0  \\)

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Then, we need to find two numbers that:

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A little thought leads us to  \\( \\var{h1} \\)  and  \\( \\var{h2} \\). these can be inserted into the brackets to give:

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\\( \\simplify{  ( x + {h1})( x + {h2} ) =0 } \\)

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Now

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If  \\(  \\simplify{ ( x + {h1}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{h1} } \\)

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If  \\(  \\simplify{ ( x + {h2}) }=0  \\)

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Then  \\(  \\simplify{ x  =-{h2} } \\)

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Factorise the following equations:

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\\( \\simplify{  x^2 + {b1}x + {c1} =0  } \\)

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 [[0]] \\( =0 \\)

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Now use these factors to give the roots of the equation:

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\\(x_1= \\)  [[1]]

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\\(x_2= \\)  [[2]]

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(If you are sure that your roots are correct but they are marked wrong, switch them around).

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\\( \\simplify{  x^2 + {b2}x + {c2} =0  } \\)

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 [[0]] \\( =0 \\)

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Now use these factors to give the roots of the equation:

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\\(x_1= \\)  [[1]]

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\\(x_2= \\)  [[2]]

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(If you are sure that your roots are correct but they are marked wrong, switch them around).

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\\( \\simplify{  x^2 + {b3}x + {c3} =0  } \\)

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 [[0]] \\( =0 \\)

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Now use these factors to give the roots of the equation:

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\\(x_1= \\) [[1]]

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\\(x_2= \\) [[2]]

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(If you are sure that your roots are correct but they are marked wrong, switch them around).

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