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# More on products

## More Products

Figure 1
Khan Academy video on products of polynomials.

We have seen how to multiply two binomials in "Product of Two Binomials". In this section, we learn how to multiply a binomial (expression with two terms) by a trinomial (expression with three terms). We can use the same methods we used to multiply two binomials to multiply a binomial and a trinomial.

For example, multiply 2x+12x+1 by x2+2x+1x2+2x+1.

( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( apply distributive law ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( expand the brackets ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( group like terms to simplify ) = 2 x 3 + 5 x 2 + 4 x + 1 ( simplify to get final answer ) ( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( apply distributive law ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( expand the brackets ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( group like terms to simplify ) = 2 x 3 + 5 x 2 + 4 x + 1 ( simplify to get final answer )
(1)

### Tip: Multiplication of Binomial with Trinomial:

If the binomial is A+BA+B and the trinomial is C+D+EC+D+E, then the very first step is to apply the distributive law:

( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E ) ( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E )
(2)

If you remember this, you will never go wrong!

### Exercise 1: Multiplication of Binomial with Trinomial

Multiply x-1x-1 with x2-2x+1x2-2x+1.

#### Solution

1. Step 1. Determine what is given and what is required :

We are given two expressions: a binomial, x-1x-1, and a trinomial, x2-2x+1x2-2x+1. We need to multiply them together.

2. Step 2. Determine how to approach the problem :

Apply the distributive law and then simplify the resulting expression.

3. Step 3. Solve the problem :
( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( expand the brackets ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( group like terms to simplify ) = x 3 - 3 x 2 + 3 x - 1 ( simplify to get final answer ) ( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( expand the brackets ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( group like terms to simplify ) = x 3 - 3 x 2 + 3 x - 1 ( simplify to get final answer )
(3)
4. Step 4. Write the final answer :

The product of x-1x-1 and x2-2x+1x2-2x+1 is x3-3x2+3x-1x3-3x2+3x-1.

### Exercise 2: Sum of Cubes

Find the product of x+yx+y


and x2-xy+y2x2-xy+y2.

#### Solution

1. Step 1. Determine what is given and what is required :

We are given two expressions: a binomial, x+yx+y, and a trinomial, x2-xy+y2x2-xy+y2.


We need to multiply them together.

2. Step 2. Determine how to approach the problem :

Apply the distributive law and then simplify the resulting expression.

3. Step 3. Solve the problem :
( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( expand the brackets ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( group like terms to simplify ) = x 3 + y 3 ( simplify to get final answer ) ( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( apply distributive law ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( expand the brackets ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( group like terms to simplify ) = x 3 + y 3 ( simplify to get final answer )
(4)
4. Step 4. Write the final answer :

The product of x+yx+y


and x2-xy+y2x2-xy+y2

is x3+y3x3+y3.

### Tip:

We have seen that:
( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3
(5)

This is known as a sum of cubes.

### Investigation : Difference of Cubes

Show that the difference of cubes (x3-y3x3-y3


) is given by the product of x-yx-y

and x2+xy+y2x2+xy+y2.

### Products

1. Find the products of:
 (a) (-2y2-4y+11)(5y-12)(-2y2-4y+11)(5y-12) (b) (-11y+3)(-10y2-7y-9)(-11y+3)(-10y2-7y-9) (c) (4y2+12y+10)(-9y2+8y+2)(4y2+12y+10)(-9y2+8y+2) (d) (7y2-6y-8)(-2y+2)(7y2-6y-8)(-2y+2) (e) (10y5+3)(-2y2-11y+2)(10y5+3)(-2y2-11y+2) (f) (-12y-3)(12y2-11y+3)(-12y-3)(12y2-11y+3) (g) (-10)(2y2+8y+3)(-10)(2y2+8y+3) (h) (2y6+3y5)(-5y-12)(2y6+3y5)(-5y-12) (i) (6y7-8y2+7)(-4y-3)(-6y2-7y-11)(6y7-8y2+7)(-4y-3)(-6y2-7y-11) (j) (-9y2+11y+2)(8y2+6y-7)(-9y2+11y+2)(8y2+6y-7) (k) (8y5+3y4+2y3)(5y+10)(12y2+6y+6)(8y5+3y4+2y3)(5y+10)(12y2+6y+6) (l) (-7y+11)(-12y+3)(-7y+11)(-12y+3) (m) (4y3+5y2-12y)(-12y-2)(7y2-9y+12)(4y3+5y2-12y)(-12y-2)(7y2-9y+12) (n) (7y+3)(7y2+3y+10)(7y+3)(7y2+3y+10) (o) (9)(8y2-2y+3)(9)(8y2-2y+3) (p) (-12y+12)(4y2-11y+11)(-12y+12)(4y2-11y+11) (q) (-6y4+11y2+3y)(10y+4)(4y-4)(-6y4+11y2+3y)(10y+4)(4y-4) (r) (-3y6-6y3)(11y-6)(10y-10)(-3y6-6y3)(11y-6)(10y-10) (s) (-11y5+11y4+11)(9y3-7y2-4y+6)(-11y5+11y4+11)(9y3-7y2-4y+6) (t) (-3y+8)(-4y3+8y2-2y+12)(-3y+8)(-4y3+8y2-2y+12)

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