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Questions asking for the factoring of simple case (a=1)
\nThese first need to be changed to standard form.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "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.
\nSome 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.
\nThe first step is to check that the equations are in \"standard form\", in these questions they are not so they will need to be re-written in standard form. Just use ordinary algebra to rearrange all terms to the left hand side \\( = 0 \\).
\nThese equations are all \"simple case\" quadratics (the co-efficient of \\( x^2 \\) is one), i.e. \\( a=1 \\).
\nWe know that, to get the \\( x^2 \\) term we need a single \\( x \\) in each bracket:
\n\\( ( x + ??)( x + ?? ) \\)
\nThen, as stated earlier, you just need to find a pair of numbers that:
\na) \\( \\simplify{ x^2 + {b1}x =-{c1} } \\)
\nThis is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.
\n\\( \\simplify{x^2 + {b1}x+{c1} }=0 \\)
\nWe know that, to get the \\( x^2 \\) term we need a single \\( x \\) in each bracket:
\n\\( ( x + ??)( x + ?? ) =0 \\)
\nThen, we need to find two numbers that:
\nA little thought leads us to \\( \\var{f1} \\) and \\( \\var{f2} \\). these can be inserted into the brackets to give:
\n\\( \\simplify{ ( x + {f1})( x + {f2} ) }=0 \\)
\n\n
\n
b) \\( \\simplify{ x^2 + {b2}x =-{c2} } \\)
\nThis is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.
\n\\( \\simplify{x^2 + {b2}x+{c2} }=0 \\)
\nWe know that, to get the \\( x^2 \\) term we need a single \\( x \\) in each bracket:
\n\\( ( x + ??)( x + ?? ) =0 \\)
\nThen, we need to find two numbers that:
\nA little thought leads us to \\( \\var{g1} \\) and \\( \\var{g2} \\). these can be inserted into the brackets to give:
\n\\( \\simplify{ ( x + {g1})( x + {g2} ) =0 } \\)
\n\n
\n
c) \\( \\simplify{ x^2 = -{c3} -{b3}x } \\)
\nThis is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.
\n\\( \\simplify{x^2 + {b3}x+{c3} }=0 \\)
\nWe know that, to get the \\( x^2 \\) term we need a single \\( x \\) in each bracket:
\n\\( ( x + ??)( x + ?? ) =0 \\)
\nThen, we need to find two numbers that:
\nA little thought leads us to \\( \\var{h1} \\) and \\( \\var{h2} \\). these can be inserted into the brackets to give:
\n\\( \\simplify{ ( x + {h1})( x + {h2} ) =0 } \\)
\n\n
\n
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Factorise the following equations:
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\nInput the equation in standard form:
\n[[0]] \\( =0 \\)
\nNow, factor the equation:
\n[[1]] \\( =0 \\)
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\nInput the equation in standard form:
\n[[0]] \\( =0 \\)
\nNow, factor the equation:
\n[[1]] \\( =0 \\)
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\nInput the equation in standard form:
\n[[0]] \\( =0 \\)
\nNow, factor the equation:
\n[[1]] \\( =0 \\)
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