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Multiplying a polynomial by a monomial is a direct application of the distributive property.
The distributive property suggests the following rule.
$10a{b}^{2}c(125{a}^{2})=1250{a}^{3}{b}^{2}c$
Determine the following products.
$3(x+8)$
$3x+24$
$(2+a)4$
$4a+8$
$({a}^{2}-2b+6)2a$
$2{a}^{3}-4ab+12a$
$8{a}^{2}{b}^{3}(2a+7b+3)$
$16{a}^{3}{b}^{3}+56{a}^{2}{b}^{4}+24{a}^{2}{b}^{3}$
$4x(2{x}^{5}+6{x}^{4}-8{x}^{3}-{x}^{2}+9x-11)$
$8{x}^{6}+24{x}^{5}-32{x}^{4}-4{x}^{3}+36{x}^{2}-44x$
$(3{a}^{2}b)(2a{b}^{2}+4{b}^{3})$
$6{a}^{3}{b}^{3}+12{a}^{2}{b}^{4}$
$5mn({m}^{2}{n}^{2}+m+{n}^{0}),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n\ne 0$
$5{m}^{3}{n}^{3}+5{m}^{2}n+5mn$
$6.03(2.11{a}^{3}+8.00{a}^{2}b)$
$12.7233{a}^{3}+48.24{a}^{2}b$
Simplify the expressions.
$(6x-1)$ .
This set of parentheses is preceded by a “
$+$ ’’ sign (implied). We simply drop the parentheses.
$(6x-1)=6x-1$
$(14{a}^{2}{b}^{3}-6{a}^{3}{b}^{2}+a{b}^{4})=14{a}^{2}{b}^{3}-6{a}^{3}{b}^{2}+a{b}^{4}$
$-(21{a}^{2}+7a-18)$ .
This set of parentheses is preceded by a “
$-$ ” sign. We can drop the parentheses as long as we change the sign of
every term inside the parentheses to its opposite sign.
$$-(21{a}^{2}+7a-18)=-21{a}^{2}-7a+18$$
$-(7{y}^{3}-2{y}^{2}+9y+1)=-7{y}^{3}+2{y}^{2}-9y-1$
Simplify by removing the parentheses.
$(2a+3b)$
$2a+3b$
$({a}^{2}-6a+10)$
${a}^{2}-6a+10$
$-(x+2y)$
$-x-2y$
$-(5m-2n)$
$-5m+2n$
$-(-3{s}^{2}-7s+9)$
$3{s}^{2}+7s-9$
Since we can consider an expression enclosed within parentheses as a single quantity, we have, by the distributive property,
For convenience we will use the commutative property of addition to write this expression so that the first two terms contain
$a$ and the second two contain
$b$ .
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