Question

Multiply:$$\begin{gathered} \left(a\right)(6x+5)(6x-5) \\ \left(b\right)(4p-7q)(4p+7q) \end{gathered}$$

Short answer

$$\begin{gathered} Multiplythebinomials,weget \\ \left(a\right)(6x+5)(6x-5) \\ 36x^2-25 \\ \left(b\right)(4p-7q)(4p+7q) \\ 16p^2-49q^2 \end{gathered}$$

Step-by-step solution

Step 1 

$$\begin{gathered} GivenInformation \\ \left(a\right)(6x+5)(6x-5) \\ \left(b\right)(4p-7q)(4p+7q) \end{gathered}$$

Step 2 

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

Part(a) Step 1. Multiply the conjugate of binomials.

$$\begin{gathered} \left(a\right) \\ Arethebinomialsconjugate(6x+5)(6x-5) \\ Itistheproductofconjuagte(a+b)(a-b) \\ Squarethefirstterm6x6x^2 \\ Squarethelastterm5{(-5)}^{2}Simplify. \\ Theprodcutisthe \\ differenceofsquares36x^2-25 \end{gathered}$$

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

$$\begin{gathered} \left(b\right) \\ Arethebinomialsconjugate(4p-7q)(4p+7q) \\ Itistheproductofconjugate(a-b)(a+b) \\ Squarethefirstterm4p{\left(4p\right)}^2 \\ Squarethelastterm7q-{\left(7q\right)}^2 \\ Simplify16p^2-49q^2 \end{gathered}$$