Solve the following problem:
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Solve the following problem:
Let's simplify the given expression by opening the parentheses using the extended distribution law:
Note that in the formula template for the above distribution law, we take as a default that the operation between the terms inside the parentheses is addition. The sign preceding the term is an inseparable part of it. We'll also apply the rules of sign multiplication and thus we can present any expression inside of the parentheses, which we'll open using the above formula, first as an expression where addition operation exists between all terms:
Let's begin by opening the parentheses:
In calculating the above multiplications, we used the multiplication table and the laws of exponents for multiplication between terms with identical bases:
In the next step, we'll combine like terms, which we define as terms where the variable(s) (or each variable separately), in this case an and b, have identical exponents. (In the absence of one of the variables from the expression, we'll consider its exponent as zero power, given that raising any number to the zero power yields 1) We'll apply the commutative law of addition as well as arrange the expression from highest to lowest power from left to right (we'll treat the free number as zero power):
In the combining of like terms performed above, we highlighted the different terms using colors, and as emphasized before, we made sure that the sign preceding the term is an inseparable part of it,
The correct answer is answer B.
\( (3+20)\times(12+4)= \)
Each term in the first binomial must multiply each term in the second binomial. With 2 terms × 2 terms, you get 4 separate products that you then combine.
Treat the sign as part of each term. So (3b + 7a)(−5a + 2b) has terms: +3b, +7a, −5a, +2b. Then apply sign rules: positive × negative = negative.
Like terms have exactly the same variables with the same exponents. For example: −15ab and +14ab are like terms, but 6b² and −35a² are not.
Yes! FOIL (First, Outer, Inner, Last) works for any two binomials. Just remember: First terms, Outer terms, Inner terms, Last terms.
That's the point! You're expanding the factored form into standard form. and are equivalent expressions.
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