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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 withthree terms). Fortunately, we use the same methods we used to multiply two binomials to multiply a binomial and a trinomial.

For example, multiply 2 x + 1 by x 2 + 2 x + 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 )
Multiplication of Binomial with Trinomial

If the binomial is A + B and the trinomial is C + 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 )

If you remember this, you will never go wrong!

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

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

  2. Apply the distributive law and then simplify the resulting expression.

  3. ( 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 )

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

Find the product of x + y and x 2 - x y + y 2 .

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

  2. Apply the distributive law and then simplify the resulting expression.

  3. ( 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. The product of x + y and x 2 - x y + y 2 is x 3 + y 3 .

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

This is known as a sum of cubes .

Investigation : difference of cubes

Show that the difference of cubes ( x 3 - y 3 ) is given by the product of x - y and x 2 + x y + y 2 .

Products

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


Factorising a quadratic

Khan academy video on factorising a quadratic.

Factorisation can be seen as the reverse of calculating the product of factors. In order to factorise a quadratic, we need to find the factors which when multiplied together equal the original quadratic.

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Source:  OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2
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