// Numbas version: finer_feedback_settings {"showstudentname": true, "question_groups": [{"pickingStrategy": "all-ordered", "pickQuestions": 1, "name": "Expanding Brackets", "questions": [{"name": "Expanding a binomial product (difference of two squares, perfect squares)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["binomial", "Binomial", "binomial product", "difference of two squares", "distributive law", "expanding", "Expanding", "factorisation", "Factorisation", "factors", "Factors", "monic", "perfect square", "quadratic"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Expand and simplify the following.

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$\\simplify{(x+{a[0]})(x-{a[0]})}$ = [[0]]

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Method 1 (difference of two squares)

Notice that the product will expand to be a difference of two squares. Therefore, we can square the first term and subtract the square of the second term (see the other methods below if you aren't sure why):

$\\simplify{(x+{a[0]})(x-{a[0]})}=\\simplify{x^2-{a[0]*a[0]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(x+{a[0]})(x-{a[0]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x-{a[0]})+{a[0]}(x-{a[0]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2-{a[0]*a[0]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{-a[0]}x$, then the Inner terms to get $\\var{a[0]}x$, and then the Last terms to get $-\\var{a[0]*a[0]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2-{a[0]*a[0]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify{(r+{a[3]})(r+{a[3]})}$ = [[0]]

Method 1 (perfect square)

Notice that $\\simplify{(r+{a[3]})(r+{a[3]})}$ is a perfect square. Therefore, we can square the first term $r$, double the product of the two terms $r$ and $\\var{a[3]}$, then square the last term $\\var{a[3]}$, add them all together (see the other methods below if you aren't sure why):

$\\simplify{(r+{a[3]})(r+{a[3]})}=\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(r+{a[3]})(r+{a[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{r(r+{a[3]})+{a[3]}(r+{a[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{r^2+{a[3]+a[3]}r+{a[3]*a[3]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $r^2$, then the Outer terms to get $\\var{a[3]}r$, then the Inner terms to get $\\var{a[3]}r$, and then the Last terms to get $\\var{a[3]*a[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{r^2+{a[3]+a[3]}r+{a[3]*a[3]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify{(x+{a[2]})^2}$ = [[0]]

It is important to realise that $\\simplify{(x+{a[2]})^2}=\\simplify[basic]{(x+{a[2]})(x+{a[2]})}$. Recall that squaring something is multiplying it by itself.

Method 1 (perfect square)

Notice that $\\simplify{(x+{a[3]})(x+{a[3]})}$ is a perfect square. Therefore, we can square the first term $r$, double the product of the two terms $r$ and $\\var{a[3]}$, then square the last term $\\var{a[3]}$, add them all together (see the other methods below if you aren't sure why):

$\\simplify{(x+{a[3]})(x+{a[3]})}=\\simplify{x^2+{2*a[3]}x+{a[3]*a[3]}}$

Method 2 (the distributive law)

We expand $\\simplify[basic]{(x+{a[3]})(x+{a[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{a[3]})+{a[3]}(x+{a[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{a[3]}x+{a[3]}x+{a[3]*a[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[3]+a[3]}x+{a[3]*a[3]}}$

Method 3 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{a[3]}x$, then the Inner terms to get $\\var{a[3]}x$, and then the Last terms to get $\\var{a[3]*a[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[3]+a[3]}x+{a[3]*a[3]}}$

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Ensure you don't use brackets in your answer.

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Expand and simplify the following.

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$\\simplify{(x+{a[0]})(x+{b[0]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(x+{a[0]})(x+{b[0]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{b[0]})+{a[0]}(x+{b[0]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[0]+b[0]}x+{a[0]*b[0]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{b[0]}x$, then the Inner terms to get $\\var{a[0]}x$, and then the Last terms to get $\\var{a[0]*b[0]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[0]+b[0]}x+{a[0]*b[0]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify{(x+{a[1]})(x+{b[1]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(x+{a[1]})(x+{b[1]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{x(x+{b[1]})+{a[1]}(x+{b[1]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$

And collect like terms: $\\simplify[basic, unitfactor]{x^2+{a[1]+b[1]}x+{a[1]*b[1]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $x^2$, then the Outer terms to get $\\var{b[1]}x$, then the Inner terms to get $\\var{a[1]}x$, and then the Last terms to get $\\var{a[1]*b[1]}$. Now add them all together: $\\simplify[basic, unitfactor]{x^2+{a[1]+b[1]}x+{a[1]*b[1]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify{(m+{a[2]})(m+{b[2]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(m+{a[2]})(m+{b[2]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{m(m+{b[2]})+{a[2]}(m+{b[2]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$

And collect like terms: $\\simplify[basic, unitfactor]{m^2+{a[2]+b[2]}m+{a[2]*b[2]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $m^2$, then the Outer terms to get $\\var{b[2]}m$, then the Inner terms to get $\\var{a[2]}m$, and then the Last terms to get $\\var{a[2]*b[2]}$. Now add them all together: $\\simplify[basic, unitfactor]{m^2+{a[2]+b[2]}m+{a[2]*b[2]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify{(t+{a[3]})(t+{b[3]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{(t+{a[3]})(t+{b[3]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{t(t+{b[3]})+{a[3]}(t+{b[3]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$

And collect like terms: $\\simplify[basic, unitfactor]{t^2+{a[3]+b[3]}t+{a[3]*b[3]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $t^2$, then the Outer terms to get $\\var{b[3]}t$, then the Inner terms to get $\\var{a[3]}t$, and then the Last terms to get $\\var{a[3]*b[3]}$. Now add them all together: $\\simplify[basic, unitfactor]{t^2+{a[3]+b[3]}t+{a[3]*b[3]}}$

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Ensure you don't use brackets in your answer.

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Expand and simplify the following.

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$\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{{c[0]}x({d[0]}x+{b[0]})+{a[0]}({d[0]}x+{b[0]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{{c[0]*d[0]}x^2+{c[0]*b[0]}x+{d[0]*a[0]}x+{a[0]*b[0]}}$

And collect like terms: $\\simplify[basic, unitfactor]{{c[0]*d[0]}x^2+{d[0]*a[0]+c[0]*b[0]}x+{a[0]*b[0]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $\\var{c[0]*d[0]}x^2$, then the Outer terms to get $\\var{c[0]*b[0]}x$, then the Inner terms to get $\\var{d[0]*a[0]}x$, and then the Last terms to get $\\var{a[0]*b[0]}$. Now add them all together: $\\simplify[basic, unitfactor]{{c[0]*d[0]}x^2+{d[0]*a[0]+c[0]*b[0]}x+{a[0]*b[0]}}$

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Ensure you don't use brackets in your answer.

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$\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ = [[0]]

Method 1 (the distributive law)

We expand $\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ one bracket at a time. Each term in the first bracket times the entire other bracket: $\\simplify[basic]{{c[1]}x({d[1]}x+{b[1]})+{a[1]}({d[1]}x+{b[1]})}$

Then we use the distributive law on each bracket: $\\simplify[basic, !collectnumbers]{{c[1]*d[1]}x^2+{c[1]*b[1]}x+{d[1]*a[1]}x+{a[1]*b[1]}}$

And collect like terms: $\\simplify[basic, unitfactor]{{c[1]*d[1]}x^2+{d[1]*a[1]+c[1]*b[1]}x+{a[1]*b[1]}}$

Method 2 (FOIL)

Multiply the First terms in each bracket to get $\\var{c[1]*d[1]}x^2$, then the Outer terms to get $\\var{c[1]*b[1]}x$, then the Inner terms to get $\\var{d[1]*a[1]}x$, and then the Last terms to get $\\var{a[1]*b[1]}$. Now add them all together: $\\simplify[basic, unitfactor]{{c[1]*d[1]}x^2+{d[1]*a[1]+c[1]*b[1]}x+{a[1]*b[1]}}$

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Ensure you don't use brackets in your answer.

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Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

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Factorise the following quadratic equations.

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Quadratic equations of the form

\\[x^2+bx+c=0\\]

can be factorised to create an equation of the form

\\[(x+m)(x+n)=0\\text{.}\\]

When we expand a factorised quadratic expression we obtain

\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]

To 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 $c$.

a)

\\[\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\\]

We need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.

\\[\\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

\\[\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\\]

b)

We can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.

\\[\\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

\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]

c)

When factorising the quadratic expression

\\[\\simplify{x^2+{v5*v6}=0}\\]

we need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.

\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}

So the factorised form of the equation is

\\[\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\\]

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$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}$

[[0]] $=0$

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

[[0]] $=0$

Your answer is not fully factorised.

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$\\simplify{x^2+{v5*v6}}=0$

[[0]] $=0$

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a)

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.

We are already given that one of them is $\\var{a}$, so we know that the other one must be $\\var{c}$.

This means our factorised equation must take the form

\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]

This expands to

\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]

So we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.

Therefore $b$ and $d$ must satisfy

\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}

$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:

\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}

So the factorised form of the equation is

\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]

b)

$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.

So the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.

", "statement": "", "variables": {"b": {"templateType": "anything", "name": "b", "description": "

$b$ in $(ax+b)(cx+d)$

", "group": "last q", "definition": "random(-5..5 except 0)"}, "c": {"templateType": "anything", "name": "c", "description": "

$c$ in $(ax+b)(cx+d)$

", "group": "last q", "definition": "random(2..8 except a)"}, "a": {"templateType": "anything", "name": "a", "description": "

$a$ in $(ax+b)(cx+d)$

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The roots of the equation

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$d$ in $(ax+b)(cx+d)$

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Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

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Factorise the equation

$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$

$(\\var{a}x+\\phantom{.}$[[0]]$) ($[[1]]$x+\\phantom{.}$[[2]]$)\\; = 0$

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Write down the roots of the equation above.

Input your answer as $x_1$ and $x_2$, where $x_1<x_2$.

$x_1=$ [[0]]

$x_2=$ [[1]]

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You have nearly spent 2 hours on this set of questions. If you are really struggling then please seek help from your Maths teacher.

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