Maths Support: Expanding brackets

Question 1

Expand the following to give a cubic in $y$.

$\simplify[std]{({a}y+{b})({c}y+{d})({p}y+{q})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answer should be a cubic in $y$ and should not include any brackets.

You can click on Show steps for more information, but you will lose one mark if you do so.

There is a video in Show steps which expands three brackets, but uses the variable $x$ rather than $y$.

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Using the method given by Show steps we first multiply out the first two brackets:

\[\begin{eqnarray*}\simplify[std]{ ({a}y+{b})({c}y+{d})}&=&\simplify[std]{{a}y*({c}y+{d})+{b}({c}y+{d})}\\&=&\simplify[std]{{a*c}y^2+{a*d}y+{b*c}y+{b*d}}\\&=&\simplify[std]{{a*c}y^2+{(a*d+b*c)}y+{b*d}}\end{eqnarray*}\]

And then multiply this by the third bracket:

\[\begin{eqnarray*}\simplify[std]{({a}y+{b})({c}y+{d})({p}y+{q})}&=&\simplify[std]{({a*c}y^2+{(a*d+b*c)}y+{b*d})({p}y+{q})}\\&=&\simplify[std]{{a*c}y^2({p}y+{q})+{(a*d+b*c)}y*({p}y+{q})+{b*d}({p}y+{q})}\\&=&\simplify[std]{{a*c*p}*y^3 +{a*c*q}*y^2+{p*(a*d+b*c)}y^2+{q*(a*d+b*c)}y+{b*d*p}y+{b*d*q}}\\&=&\simplify[std]{{a*c*p}y^3+{a*c*q+a*d*p+p*b*c}y^2+{a*d*q+b*c*q+b*d*p}y+{b*d*q}}\end{eqnarray*}\]

Question 2

Expand the following to give quadratics in $x$.

$\simplify[std]{({a}x)({c}x+{d})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

$\simplify[std]{({a1}x+{b1})({c1}x)}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answers should be quadratics in $x$ and should not include any brackets.

You can click on Show steps to get more information, but you will lose one mark if you do so.

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1. Using the method given by Show steps we have:

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

\[\simplify[std]{ ({a1}x+{b1})*({c1}x)}=\simplify[std]{{a1*c1}x^2+{b1*c1}x}\]

Question 3

Expand the following to give a quadratic in $x$.

$\simplify[std]{({a}x+{b})({c}x+{d})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answer should be a quadratic in $x$ and should not include any brackets.

You can click on Show steps for more information, but you will lose one mark if you do so.

(You will lose 1 mark.)

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Using the method given by Show steps we have:

\[\begin{eqnarray*}\simplify[std]{ ({a}x+{b})({c}x+{d})}&=&\simplify[std]{{a}x*({c}x+{d})+{b}({c}x+{d})}\\&=&\simplify[std]{{a*c}x^2+{a*d}x+{b*c}x+{b*d}}\\&=&\simplify[std]{{a*c}x^2+{(a*d+b*c)}x+{b*d}}\end{eqnarray*}\]

Question 4

Expand the following to give a cubic in $w$.

$\simplify[std]{({p}w+{q})({a}w^2+{b}w+{c})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answer should be a cubic in $w$ and should not include any brackets.

You can click on Show steps for more information, but you will lose one mark if you do so.

(You will lose 1 mark.)

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Using the method given by Show steps:

\[\begin{eqnarray*}\simplify[std]{ ({p}w+{q})({a}w^2+{b}w+{c})}&=&\simplify[std]{{p}w*({a}w^2+{b}w+{c})+{q}({a}w^2+{b}w+{c})}\\&=&\simplify[std]{{a*p}w^3+{b*p}w^2+{c*p}w+{a*q}w^2+{q*b}w+{c*q}}\\&=&\simplify[std]{{a*p}w^3+{(a*q+b*p)}w^2+{b*q+c*p}w+{c*q}}\end{eqnarray*}\]

Question 5

Expand the following to give a cubic in $z$.

$\simplify[std]{({a}z^2+{b}z+{c})({p}z+{q})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answer should be a cubic in $z$ and should not include any brackets.

You can click on Show steps for more information, but you will lose one mark if you do so.

(You will lose 1 mark.)

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Using the method given by Show steps:

\[\begin{eqnarray*}\simplify[std]{ ({a}z^2+{b}z+{c})({p}z+{q})}&=&\simplify[std]{{a}z^2*({p}z+{q})+{b}*z*({p}z+{d})+{c}({p}z+{q})}\\&=&\simplify[std]{{a*p}z^3+{a*q}z^2+{b*p}z^2+{b*d}z+{c*p}z+{c*q}}\\&=&\simplify[std]{{a*p}z^3+{(a*q+b*p)}z^2+{b*d+c*p}z+{c*q}}\end{eqnarray*}\]

Question 6

Expand the following to give a quartic in $z$.

$\simplify[std]{({a}z^2+{b}z+{c})({m}*z^2+{p}z+{q})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer: .

Your answer should be a quartic (degree 4 polynomial) in $z$ and should not include any brackets.

You can click on Show steps for more information, but you will lose one mark if you do so.

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Using the method given by Show steps:

\[\begin{eqnarray*}\simplify[std]{ ({a}z^2+{b}z+{c})({m}z^2+{p}z+{q})}&=&\simplify[std]{{a}z^2*({m}z^2+{p}z+{q})+{b}*z*({m}z^2+{p}z+{q})+{c}({m}z^2+{p}z+{q})}\\&=&\simplify[std]{{a*m}z^4+{a*p}z^3+{a*q}z^2+{b*m}z^3+{b*p}z^2+{b*q}z+{c*m}z^2+{c*p}z+{c*q}}\\&=&\simplify[std]{{a*m}z^4+{a*p+m*b}z^3+{(a*q+c*m+b*p)}z^2+{b*q+c*p}z+{c*q}}\end{eqnarray*}\]

Question 7

Simplify the following expression.

Simplify:

$\simplify[std]{({a}x+{b})({c}x+{d})-({a}x+{d})({c}x+{b})}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Do not include brackets in your answer.

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Expanding the brackets we have:

\[\begin{eqnarray*}\simplify[std]{({a}x+{b})({c}x+{d})-({a}x+{d})({c}x+{b})}&=&(\simplify[std]{{a*c}x^2+{b*c+a*d}x+{b*d}})-(\simplify[std]{{a*c}x^2+{b*a+c*d}x+{b*d}})\\&=&\simplify[std]{{b*c+a*d}x-{b*a+c*d}x}\\&=&\var{(a-c)*(d-b)}x\end{eqnarray*}\]

Question 8

Simplify the following expression.

Simplify:

$\simplify[std]{({a}x+{b}y)({c}x+{d}y)-({a}x+{d}y)({c}x+{b}y)}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Do not include brackets in your answer.

Input $xy$ as $x*y$.

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Expanding the brackets we have:

\[\begin{eqnarray*}\simplify[std]{({a}x+{b}y)({c}x+{d}y)-({a}x+{d}y)({c}x+{b}y)}&=&(\simplify[std]{{a*c}x^2+{b*c+a*d}x*y+{b*d}y^2})-(\simplify[std]{{a*c}x^2+{b*a+c*d}x*y+{b*d}y^2})\\&=&\simplify[std]{{b*c+a*d}x*y-{b*a+c*d}x*y}\\&=&\var{(a-c)*(d-b)}xy\end{eqnarray*}\]

Question 9

Simplify the following expression.

Simplify:

$\simplify[std]{({a}x+{b}y)({c}x+{d}y)-({a}x+{d}y)({c}x+{b}y)}=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Do not include brackets in your answer.

Input $xy$ as $x*y$.

This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback.

Score: 0/2

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Expanding the brackets we have:

Question 10

Express the following expression as $ax+by$ for suitable integers $a$ and $b$.

Simplify $f(x,y)=\simplify[std]{{a}({b}x+{c}y)+{a1}x+{b1}y+{a2}({b2}x+{c2}y)}$

$f(x,y)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Do not include brackets in your answer.

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\[\begin{eqnarray*}f(x,y)&=&\simplify[std]{{a}({b}x+{c}y)+{a1}x+{b1}y+{a2}({b2}x+{c2}y)}\\&=&\simplify[std]{{a*b}x+{a*c}y+{a1}x+{b1}y+{a2*b2}x+{a2*c2}y}\\&=&\simplify[std]{{a*b}x+{a1}x+{a2*b2}x+{a*c}y+{b1}y+{a2*c2}y}\\&=&\simplify[std]{({a*b}+{a1}+{a2*b2})x+({a*c}+{b1}+{a2*c2})y}\\&=&\simplify[std]{{a*b+a1+a2*b2}x+{a*c+b1+a2*c2}y}\end{eqnarray*}\]

Question 11

Express the following expression as $ax+by$ for suitable integers $a$ and $b$.

Simplify $f(x,y)=\simplify[std]{{a}({b}x+{c}y)+{a1}x+{b1}y+{a2}({b2}x+{c2}y)}$

$f(x,y)=\;$ $\displaystyle{}$ question.score feedback.none Expected answer:

Do not include brackets in your answer.

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Score: 0/1

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Maths Support: Expanding brackets

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