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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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