Answer:
c. (2x + 3y)² = (2x)² + 2(2x)(3y) + (<u>3y</u>)²
d. (w - 1)(1 + w + w²) = <u> </u><u>w³</u> - 1
e. a² - 16 = (a + <u>4</u>)(a - <u>4</u>)
f. (2x + 5y)(2x - 5y) = <u>4x²</u> - <u>25y²</u>
g. (2²¹ - 1)(2²¹ + 1) = <u>2⁴²</u> - 1
h. [(x - y) - 3][(x - y) + 3] = (<u>x - y</u>)² - 9
Step-by-step explanation:
c. Although this expression, when squared, is really <em>4x</em><em>²</em><em> </em><em>+</em><em> </em><em>12xy</em><em> </em><em>+</em><em> </em><em>9y</em><em>²,</em><em> </em>they just splitted it up into several more terms. In this last blank should be a 9y², but instead, they took the square root of the term, which is <em>3y,</em><em> </em>and set the square outside of the parentheses.
*I bet that was what made it confusing, but have no fear. The mathematics tutor is here.
d. This is similar to the Quadratic Equation [<em>y</em><em> </em><em>=</em><em> </em><em>Ax</em><em>²</em><em> </em><em>+</em><em> </em><em>Bx</em><em> </em><em>+</em><em> </em><em>C</em>] where you can, what I like to call... FOIL it, however, you have an extended amount of terms. Although you still have to FOIL it, it is just a unique procedure. This time, you have to one-by-one, take each term in the first set of parentheses [-1 and w] and multiply them by the second all three terms in the second set of parentheses.
e. This is a piece of cake because we know BOTH square roots of 16 and we are multiplying what are called conjugates.
f. Again, we are multiplying conjugates, so we just multiply the first and last two terms in each set of parentheses.
g. Again, we are multiplying conjugates EXCEPT now, we have exponents in each set of parentheses. Whenever you multiply terms, you add the exponents and your base remains the same, which is what you see in the above answer.
h. For the final exercise, we are multiplying conjugates again, and we obviously can see what fills that blank because there are TWO IDENTICAL EXPRESSIONS that are being multiplied, which in this case are being "squared", and that expression is <em>x</em><em> </em><em>-</em><em> </em><em>y</em><em>.</em><em> </em>It appears twice, so that is a giveaway.
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**Conjugates are IDENTICAL expressions with OPPOSITE operation symbols. Whenever you multiply conjugates, just add the first and last two terms in both sets of parentheses because when you FOIL, the midst term will ALWAYS result in zero.
