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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 5: Q.441R (page 561)

Multiply the binomials by using
(a) The Distributive property (b) the FOIL method (c) the Vertical method.
$(6 p - 11) (3 p - 10)$ .

Short Answer

Expert verified

(a) Using the distributive property $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

(b) using the FOIL method $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

(c) Using the Vertical method $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

Step by step solution

01

Step 1

(a) Use the distributive property $a 路 (b + c) = a 路 b + a 路 c$ as follows:

$\begin{array}{rcl} (6 p - 11) (3 p - 10) & = & 6 p (3 p - 10) - 11 (3 p - 10) \\ = & 6 p (3 p) + (6 p) (- 10) + (- 11) (3 p) + (- 11) (- 10) \\ = & 18 p^{2} - 60 p - 33 p + 110 \\ = & 18 p^{2} - 93 p + 110 \end{array}$

So, $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

02

Step 2

(b) Use the FOIL method to multiply the binomials $(6 p - 11) (3 p - 10)$ .

Multiply First terms $6 p, 3 p$ : $6 p 路 3 p = 18 p^{2}$ .

Multiply Outer terms $6 p, - 10$ : $6 p (- 10) = - 60 p$ .

Multiply Inner terms $- 11, 3 p$ : $(- 11) (3 p) = - 33 p$ .

Multiply Last terms $- 11, - 10$ : $(- 11) (- 10) = 110$ .

Finally, add the terms:

$\begin{array}{rcl} (6 p - 11) (3 p - 10) & = & 18 p^{2} - 60 p - 33 p + 110 \\ = & 18 p^{2} - 93 p + 110 \end{array}$

So, $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

03

Step 3

(c) Use the vertical method.

$6 p - 11 脳 3 p - 10 - 60 p + 110 18 p^{2} - 33 p 18 p^{2} - 93 p + 110$

So, $(6 p - 11) (3 p - 10) = 18 p^{2} - 93 p + 110$ .

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