Q5.66 - Intermediate Algebra

Question

Multiply: $$\begin{gathered} \left(a\right) (2x+7)(2x-7) \\ \left(b\right) (3x-y)(3x+y) \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} Multiple the binomials , we get \\ \left(a\right) (2x+7)(2x-7) \\ 4x^2-49 \\ \left(b\right) (3x-y)(3x+y) \\ 9x^2-y^2 \end{gathered}$$

Step 1

$$\begin{gathered} Given Information \\ \left(a\right) (2x+7)(2x-7) \\ \left(b\right) (3x-y)(3x+y) \end{gathered}$$

Step 2

$$\begin{gathered} A binomial conjugate is found simply by changing the sign of the 2nd term. \\ The binomial conjugate of (a+b) is (a-b) \end{gathered}$$

Part(a) Step1. Multiply the conjugates of binomials.

$$\begin{gathered} \left(a\right) \\ Are the binomials conjugate (2x+7)(2x-7) \\ It is the products of conjugate (a+b)(a-b) \\ Square the first term {\left(2x\right)}^2 \\ Square the last term -{\left(7\right)}^2 \\ Simplify 4x^2-49 \end{gathered}$$

Part(b) Step1. Multiply the conjugates of binomials.

$$\begin{gathered} \left(b\right) \\ Are the binomials conjugates? (3x-y)(3x+y) \\ It is the product of conjugate (a-b)(a+b) \\ Square of the first term 3x {\left(3x\right)}^2 \\ Square the last term y -{\left(y\right)}^2 \\ Simplify 9x^2-y^2 \end{gathered}$$