// Numbas version: finer_feedback_settings {"name": "Quadratic Equations: Factorise, Simple Case (Instructional)", "metadata": {"description": "

Structured questions with instructions.

Use for coaching not assessment.

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

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

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.

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.

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 \\).

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

\\( ( x + ??)( x + ?? ) \\)

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

Multiply to give \\( c \\)
Add to give \\(b \\).

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c1} \\)
- Add to give \\( \\var{b1} \\).
\n

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

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

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c2} \\)
- Add to give \\( \\var{b2} \\).
\n

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

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

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c3} \\)
- Add to give \\( \\var{b3} \\).
\n

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

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

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Factorise the following equations:

"}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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

[[0]] \\( =0 \\)

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

[[0]] \\( =0 \\)

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

[[0]] \\( =0 \\)

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

These first need to be changed to standard form.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

", "advice": "

We are asked to factor different quadratic equations.

The 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 \\).

These equations are all \"simple case\" quadratics (the co-efficient of \\( x^2 \\) is one), i.e. \\( a=1 \\).

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

\\( ( x + ??)( x + ?? ) \\)

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

Multiply to give \\( c \\)
Add to give \\(b \\).

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c1} \\)
- Add to give \\( \\var{b1} \\).
\n

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

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

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c2} \\)
- Add to give \\( \\var{b2} \\).
\n

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

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

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c3} \\)
- Add to give \\( \\var{b3} \\).
\n

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

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

Factorise the following equations:

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

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

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

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

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

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

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

Then using factors to find roots.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

", "advice": "

We are asked to factor different quadratic equations.

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

\\( ( x + ??)( x + ?? ) \\)

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

Multiply to give \\( c \\)
Add to give \\(b \\).

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

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

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c1} \\)
- Add to give \\( \\var{b1} \\).
\n

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

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

Now

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

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

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

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

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c2} \\)
- Add to give \\( \\var{b2} \\).
\n

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

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

Now

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

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

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

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

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c3} \\)
- Add to give \\( \\var{b3} \\).
\n

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

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

Now

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

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

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

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

Factorise the following equations:

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

[[0]] \\( =0 \\)

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

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

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

_{(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 } \\)

[[0]] \\( =0 \\)

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

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

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

_{(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 } \\)

[[0]] \\( =0 \\)

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

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

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

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

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

These first need to be changed to standard form.

Then use factors to solve.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

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

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

", "advice": "

We are asked to factor different quadratic equations.

These equations are all \"simple case\" quadratics (the co-efficient of \\( x^2 \\) is one), i.e. \\( a=1 \\).

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

\\( ( x + ??)( x + ?? ) \\)

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

Multiply to give \\( c \\)
Add to give \\(b \\).

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c1} \\)
- Add to give \\( \\var{b1} \\).
\n

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

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

Now

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

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

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

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

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c2} \\)
- Add to give \\( \\var{b2} \\).
\n

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

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

Now

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

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

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

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

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

This is not \"standard\". Re-arrange all terms to the left hand side, leaving \\( 0 \\) on the right.

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

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

\\( ( x + ??)( x + ?? ) =0 \\)

Then, we need to find two numbers that:

\n
- Multiply to give \\( \\var{c3} \\)
- Add to give \\( \\var{b3} \\).
\n

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

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

Now

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

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

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

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

Factorise the following equations:

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

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

Use these factors to give the roots of the equation:

\\(x_1= \\) [[2]]

\\(x_2= \\) [[3]]

_{(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} } \\)

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

Use these factors to give the roots of the equation:

\\(x_1= \\) [[2]]

\\(x_2= \\) [[3]]

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

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

Input the equation in standard form:

[[0]] \\( =0 \\)

Now, factor the equation:

[[1]] \\( =0 \\)

Use these factors to give the roots of the equation:

\\(x_1= \\) [[2]]

\\(x_2= \\) [[3]]

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

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