Mckenzie Ramos
11/24/2023 · Primary School
5. Multiplica cada trinomio de la forma \( a x^{2 n}+b x^{n}+c \) por el valor de \( a \) \( \begin{array}{ll}\text { a. } 3 x^{2}+6 x+2 & \text { b. } 2 x^{2}-5 x+4 \\ \text { c. } 6 x^{2}-10 x+7 & \text { d. } 5 x^{4}+3 x^{2}+1 \\ \text { e. } 7 x^{5}-4 x^{4}+3 & \text { f. } 4 x^{2}-x+3\end{array} \)
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Step-by-step Solution
Para multiplicar cada trinomio de la forma \( a x^{2 n}+b x^{n}+c \) por el valor de \( a \), simplemente multiplicamos cada término del trinomio por el coeficiente \( a \). A continuación, se presentan los resultados:
a. \( 3 x^{2}+6 x+2 \)
Multiplicamos por \( a = 3 \):
\[ 3(3 x^{2}) + 3(6 x) + 3(2) = 9 x^{2} + 18 x + 6 \]
b. \( 2 x^{2}-5 x+4 \)
Multiplicamos por \( a = 2 \):
\[ 2(2 x^{2}) + 2(-5 x) + 2(4) = 4 x^{2} - 10 x + 8 \]
c. \( 6 x^{2}-10 x+7 \)
Multiplicamos por \( a = 6 \):
\[ 6(6 x^{2}) + 6(-10 x) + 6(7) = 36 x^{2} - 60 x + 42 \]
d. \( 5 x^{4}+3 x^{2}+1 \)
Multiplicamos por \( a = 5 \):
\[ 5(5 x^{4}) + 5(3 x^{2}) + 5(1) = 25 x^{4} + 15 x^{2} + 5 \]
e. \( 7 x^{5}-4 x^{4}+3 \)
Multiplicamos por \( a = 7 \):
\[ 7(7 x^{5}) + 7(-4 x^{4}) + 7(3) = 49 x^{5} - 28 x^{4} + 21 \]
f. \( 4 x^{2}-x+3 \)
Multiplicamos por \( a = 4 \):
\[ 4(4 x^{2}) + 4(-x) + 4(3) = 16 x^{2} - 4 x + 12 \]
Así que los resultados finales son:
a. \( 9 x^{2} + 18 x + 6 \)
b. \( 4 x^{2} - 10 x + 8 \)
c. \( 36 x^{2} - 60 x + 42 \)
d. \( 25 x^{4} + 15 x^{2} + 5 \)
e. \( 49 x^{5} - 28 x^{4} + 21 \)
f. \( 16 x^{2} - 4 x + 12 \)
Quick Answer
a. \( 9 x^{2} + 18 x + 6 \)
b. \( 4 x^{2} - 10 x + 8 \)
c. \( 36 x^{2} - 60 x + 42 \)
d. \( 25 x^{4} + 15 x^{2} + 5 \)
e. \( 49 x^{5} - 28 x^{4} + 21 \)
f. \( 16 x^{2} - 4 x + 12 \)
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