### What makes these products "special"?

On this page, you will find algebraic products you will use repeatedly in this chapter and in a lot of math you will encounter later on.They are "special" because they are very common, and they are worth knowing about.

It is easier later on if you can recognize these products easily.

## Special Products involving Squares

Multiplying out brackets creates the following special products.Knowing these well is worth knowing because you'll need them often.

*a*(*x *+* y*) = *ax *+* ay * (Distributive Law)

(*x* + *y*)(*x* − *y*) = *x*2 − *y*2 (Difference of 2 squares)

(*x* + *y*)2 = *x*2 + 2*xy *+ *y*2 (Square of a sum)

(*x* − *y*)2 = *x*2 − 2*xy *+ *y*2 (Square of a difference)

### Examples using the special products

In Example 1, multiply out 2*(a − 3)

Provide an answer

This product utilizes the first one above.We multiply the term outside the bracket (the "2x") by the terms within the brackets (the "a" and "−3").

2*x*(*a* − 3) = 2*ax* − 6*x*

We recognize this one involves the **Difference of 2 squares**:

(7*s* + 2*t*)(7*s* − 2*t*)

= (7*s*)2 − (2*t*)2

= 49*s*2 − 4*t*2

(12* *+ 5*ab*)(12 − 5*ab*)

= (12)2 − (5*ab*)2

= 144 − 25*a*2*b*2

The answer is a difference of 2 squares.

This one is the **square of a sum of 2 terms**.

(5*a* + 2*b*)2

= (5*a*)2 + 2(5*a*)(2*b*) + (2*b*)2

= 25*a*2 + 20*a**b* + 4*b*2

(q − 6)2

= (q)2 − 2(*q*)(6) + (6)2

= q2 − 12*q *+ 36

This example involved the square of a difference of 2 terms.

This question is not in any of the formats we have above. So we just need to multiply out the brackets, term-by-term.

It"s important to recognize when we have a Special Product and when our question is something else.

(8*x* −* y*)(3*x* + 4*y*)

= 8*x*(3*x* + 4*y*) −* y*(3*x* + 4*y*)

= 8*x*(3*x*) + 8*x*(4*y*) −* y*(3*x*) −* y*(4*y*)

= 24*x*2 + 32*x**y* −3*xy *− 4*y*2

= 24*x*2 + 29*x**y* − 4*y*2

To expand this, we put it in the form (*a* + *b*)2 and expand it using the third rule above, which says:

(*a* + *b*)2 = *a*2 + 2*ab* + *b*2

I put

*a* = *x* + 2

*b* = 3*y*

This gives me:

(*x* + 2 + 3*y*)2

*b*)2 step.>

= (<*x* + 2> + 3*y*)2

= <*x* + 2>2 + 2<*x* + 2>(3*y*) + (3*y*)2

*b*)2 = *a*2 + 2*ab* + *b*2>

= <*x*2 + 4*x* + 4> + (2*x* + 4)(3*y*) + 9*y*2

= *x*2 + 4*x* + 4 + 6*xy* + 12*y* + 9*y*2

I could have chosen the following and obtained the same answer:

*a* = *x*

*b* = 2 + 3*y*

Try it!

## Special Products involving Cubes

The following products are just the result of multiplying out the brackets.

(*x* + *y*)3 = *x*3 + 3*x*2*y *+ 3*xy*2 + *y*3 (Cube of a sum)

(*x* −* y*)3 = *x*3 − 3*x*2*y *+ 3*xy*2 − *y*3 (Cube of a difference)

(*x* + *y*)(*x*2 − *xy *+ *y*2) = *x*3 + *y*3 (Sum of 2 cubes)

(*x* − *y*)(*x*2 + *xy *+ *y*2) = *x*3 − *y*3 (Difference of 2 cubes)

These are also worth knowing well enough so you recognize the form, and the differences between each of them. (Why? Because it"s easier than multiplying out the brackets and it helps us solve more complex algebra problems later.)

**Example 8: ** Expand(2*s* + 3)3

Answer

This involves the Cube of a Sum:

(2*s* + 3)3

= (2*s*)3 + 3(2*s*)2(3) + 3(2*s*)(3)2 + (3)3

= 8*s*3 + 36*s*2* + * 54*s *+ 27

### Exercises

Expand:

(1) (s + 2t)(s − 2t)

Answer

Using the Difference of 2 Squares formula

(*x* + *y*)(*x* − *y*) = *x*2 − *y*2,

we have:

(s + 2t)(s − 2t)

= (s)2 − (2t)2

= s2 − 4t2

(2) (i1 + 3)2

Answer

Using the Square of a Sum formula

(*x* + *y*)2 = *x*2 + 2*xy *+ *y*2,

we have:

(i1 + 3)2

= (i1)2 + 2(i1)(3) + (3)2

= i12 + 6i1 + 9

(3) (3x + 10y)2

Answer

Using the Square of a Sum formula

(*x* + *y*)2 = *x*2 + 2*xy *+ *y*2,

we have:

(3x + 10y)2

= (3x)2 + (2)(3x)(10y) + (10y)2

= 9x2 + 60xy + 100y2

(4) (3p − 4q)2

Answer

Using the Square of a Difference formula

(*x* − *y*)2 = *x*2 − 2*xy *+ *y*2,

we have:

(3p − 4q)2

= (3p)2 − (2)(3p)(4q) + (4q)2

= 9p2 − 24pq + 16q2

top

Factoring and Fractions

2. Common Factor and Difference of Squares

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