OpenStax-CNX module: m39387
1
Products and factors: More on
products
∗
Free High School Science Texts Project
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
1 More Products
Khan Academy video on products of polynomials.
This media object is a Flash object. Please view or download it at
<http://www.youtube.com/v/fGThIRpWEE4&rel=0>
Figure 1
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 + 1 by x2 + 2x + 1.
(2x + 1) x2 + 2x + 1
=
2x x2 + 2x + 1 + 1 x2 + 2x + 1
= 2x x2 + 2x (2x) + 2x (1) + 1 x2 + 1 (2x) + 1 (1)
=
=
2x3 + 4x2 + 2x + x2 + 2x + 1
2x3 + 4x2 + x2 + (2x + 2x) + 1
3
=
2
2x + 5x + 4x + 1
(apply distributive law)
(1)
(expand the brackets)
(group like terms to simplify)
(simplify to get final answer)
tip: If the binomial is A + B and the trinomial is C + D + E , then the very rst 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!
∗ Version
1.1: Aug 2, 2011 2:12 am -0500
† http://creativecommons.org/licenses/by/3.0/
http://cnx.org/content/m39387/1.1/
(2)
OpenStax-CNX module: m39387
2
Exercise 1: Multiplication of Binomial with Trinomial
(Solution on p. 3.)
Exercise 2: Sum of Cubes
(Solution on p. 3.)
Multiply x − 1 with x2 − 2x + 1.
Find the product of x + y and x2 − xy + y 2 .
tip:
We have seen that:
(3)
(x + y) x2 − xy + y 2 = x3 + y 3
This is known as a sum of cubes.
1.1 Investigation : Dierence of Cubes
Show that the dierence of cubes (x3 − y 3 ) is given by the product of x − y and x2 + xy + y 2 .
1.2 Products
1. Find the products of:
(a) −2y 2 − 4y + 11 (5y − 12)
(c) 4y 2 + 12y + 10 −9y 2 + 8y + 2
(e) 10y 5 + 3 −2y 2 − 11y + 2
(g) (−10) 2y 2 + 8y + 3
(i) 6y 7 − 8y 2 + 7 (−4y − 3) −6y 2 − 7y − 11
(k) 8y 5 + 3y 4 + 2y 3 (5y + 10) 12y 2 + 6y + 6
(m) 4y 3 + 5y 2 − 12y (−12y − 2) 7y 2 − 9y + 12
(o) (9) 8y 2 − 2y + 3
(q) −6y 4 + 11y 2 + 3y (10y + 4) (4y − 4)
(s) −11y 5 + 11y 4 + 11 9y 3 − 7y 2 − 4y + 6
Table 1
Click here for the solution1
1 http://www.fhsst.org/llz
http://cnx.org/content/m39387/1.1/
(b) (−11y + 3) −10y 2 − 7y − 9
(d) 7y 2 − 6y − 8 (−2y + 2)
(f) (−12y − 3) 12y 2 − 11y + 3
(h) 2y 6 + 3y 5 (−5y − 12)
(j) −9y 2 + 11y + 2 8y 2 + 6y − 7
(l) (−7y + 11) (−12y + 3)
(n) (7y + 3) 7y 2 + 3y + 10
(p) (−12y + 12) 4y 2 − 11y + 11
(r) −3y 6 − 6y 3 (11y − 6) (10y − 10)
(t) (−3y + 8) −4y 3 + 8y 2 − 2y + 12
OpenStax-CNX module: m39387
3
Solutions to Exercises in this Module
Solution to Exercise (p. 2)
Step 1. We are given two expressions: a binomial, x − 1, and a trinomial, x2 − 2x + 1. We need to multiply
them together.
Step 2. Apply the distributive law and then simplify the resulting expression.
Step 3.
(x − 1) x2 − 2x + 1
=
x x2 − 2x + 1 − 1 x2 − 2x + 1
= x x2 + x (−2x) + x (1) + −1 x2 − 1 (−2x) − 1 (1)
(apply distributive law)
=
x3 − 2x2 + x − x2 + 2x − 1
(expand the brackets)
=
x3 + −2x2 − x2 + (x + 2x) − 1
(group like terms to simplify)
=
x3 − 3x2 + 3x − 1
(simplify to get final answer)
(4)
Step 4. The product of x − 1 and x2 − 2x + 1 is x3 − 3x2 + 3x − 1.
Solution to Exercise (p. 2)
Step 1. We are given two expressions: a binomial, x + y , and a trinomial, x2 − xy + y 2 . We need to multiply
them together.
Step 2. Apply the distributive law and then simplify the resulting expression.
Step 3.
(x + y) x2 − xy + y 2
=
x x2 − xy + y 2 + y x2 − xy + y 2
= x x2 + x (−xy) + x y 2 + y x2 + y (−xy) + y y 2
(apply distributive law)
=
x3 − x2 y + xy 2 + yx2 − xy 2 + y 3
(expand the brackets)
=
x3 + −x2 y + yx2 + xy 2 − xy 2 + y 3
(group like terms to simplify)
=
x3 + y 3
(simplify to get final answer)
Step 4. The product of x + y and x2 − xy + y 2 is x3 + y 3 .
http://cnx.org/content/m39387/1.1/
(5)