Respuesta :

r3t40

The distribution is very simple. Using FOIL.

It states that [tex](a+b)(c+d)=ac+ad+bc+bd[/tex].

Also note that when multiplying expressions we multiply variables and values differently. If we have variables like [tex]x[/tex] their exponents will add. If we have values like 3 we multiply them normally.

For example your practise 3.

[tex](x+3)(2x^2+4)=2x^2\cdot x+4x+3\cdot2x^2+3\cdot4 \\

\underline{2x^3+4x+6x^2+12}

[/tex]

Now just order the expressions from bigger exponent to smaller and than values. (Usual notation although no need).

And solution is:

[tex]\boxed{2x^3+6x^2+4x+12}[/tex]

Hope this helps.

r3t40

znk

Answer:

See below

Step-by-step explanation:

[tex]a\cdot(b + c) = a\cdot b + a\cdot c[/tex]

3) Practice: Organizing information

[tex]\begin{array}{lll}\qquad \textbf{Steps} & \textbf{Problem: }(x + 3)(2x^{2} + 4) & \\\textbf{1. List variables} & a = x + 3 & \\ & b = 2x^{2} & \\ & c = 4 &\\\\\textbf{2. Distribute} & (x + 3)(2x^{2} + 4)& = (x + 3)(2x^{2}) + (x + 3)(4)\\\\\textbf{3. Redistribute} & (x + 3)(2x^{2})& (x + 3)(4)\\& a = 2x^{2} & a = 4\\& b = x & b = x\\& c = 3 & c = 3\\& 2x^{3} + 6x^{2} & 4x + 12\\\textbf{4. Combine}& & \\\qquad\textbf{terms} & 2x^{3} + 6x^{2}+ 4x + 12 & \\\end{array}[/tex]

4. Practice: Summarizing

[tex]\text{You can use the FOIL method to multiply two }\underline{\text{binomials}}.\\\text{The letters in FOIL stand for }\underline{\text{First, Outer, Inner, Last}}.\\\text{The FOIL method helps you to remember how to multiply each term in one }\\\underline{\text{binomial}} \text{ by each term in the other }\underline{\text{binomial}}.[/tex]