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Musya8 [376]
1 year ago
12

Caden states that n^2 +3n + 2n is an equivalent expression to 6n. Why is Caden's statement incorrect?

Mathematics
1 answer:
malfutka [58]1 year ago
7 0
\begin{gathered} \text{The simplification of n}^2+3n+2n\text{ is:} \\ i)n^2+5n \\ By\text{ collecting common term, this can be written in form of:} \\ ii)\text{ n(n+5)} \end{gathered}

Thus, options A and D hold, from the simplifications above.

Let's consider the validity of the remaining options provided.

\begin{gathered} \text{For option B)} \\ \text{substitute for n=1 into the expression n}^2+3n+2n,\text{ we have} \\ 1^2+3(1)+2(1)=1+3+2=6 \\ \text{substitute for n=1 into the expression 6n, we have} \\ 6(1)=6 \\ \text{Thus, the expression n}^2+3n+2n\text{ is equivalent to 6n, for n=1} \end{gathered}\begin{gathered} \text{For option C)} \\ \text{The expression n}^2+3n+2n\text{ does not simplify to 7n} \end{gathered}\begin{gathered} \text{For option E)} \\ \text{substitute for n=4 into the expression n}^2+3n+2n,\text{ we have:} \\ 4^2+3(4)+2(4)=16+12+8=36 \\ \text{substitute for n=6 into the expression 6n, we have:} \\ 6(4)=24 \\ \text{Thus, the two(2) expressions are not equivalent to each other, for n=4} \end{gathered}\begin{gathered} \text{For option F)} \\ \text{substitute for n=3 into the expression n}^2+3n+2n,\text{ we have:} \\ 3^2+3(3)+2(3)=9+9+6=24 \\ \text{substitute for n=3 into the expression 6n, we have:} \\ 6(3)=18 \\ \text{Thus, the two(2) expressions are not equivalent to each other, for n=3} \end{gathered}

Hence, the correct options that apply are options A, D, E and F

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Jay factored the 4 term polynomial : x^3 - 9x +2x^2 - 18 and decided that the complete factorization was : ( x + 2 ) (x^2-9 ). B
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Answer:

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Step-by-step explanation:

we are given that x^3-9x+2x^2-18

w are sked to step by step factorise the above polynomial

\left(x^3+2x^2\right)+\left(-9x-18\right)

-9\mathrm{\:from\:}-9x-18\mathrm{:\quad }-9\left(x+2\right)

-9x-9\cdot \:2

-9\left(x+2\right)

\mathrm{Factor\:out\:}x^2\mathrm{\:from\:}x^3+2x^2\mathrm{:\quad }x^2\left(x+2\right)

-9\left(x+2\right)+x^2\left(x+2\right)

\left(x+2\right)\left(x^2-9\right)

x^2-9:\quad \left(x+3\right)\left(x-3\right)

x^2-9

\mathrm{Rewrite\:}9\mathrm{\:as\:}3^2

=x^2-3^2

\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)

x^2-3^2=\left(x+3\right)\left(x-3\right)

Hence

x^3-9x+2x^2-18=\left(x+2\right)\left(x+3\right)\left(x-3\right)

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Answer:

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Step-by-step explanation:

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Answer:

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