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3 years ago

What is the standard form polynomial that represents this product

Mathematics

1 answer:

NISA [10]3 years ago

5 0

Answer:

(-8m⁵ + 10m⁴ + 9m³- 16m² + 5m)

Step-by-step explanation:

Given polynomial is (-2m³ + 3m² - m)(4m² + m - 5).

To get the standard form of this polynomial we will simplify the given expression.

(-2m³ + 3m² - m)(4m² + m - 5)

= -2m³(4m² + m - 5) + 3m²(4m² + m - 5) - m(4m² + m - 5)

= -8m⁵- 2m⁴ + 10m³+ 12m⁴ + 3m³ - 15m² - 4m³ - m² + 5m

= -8m⁵ + 10m⁴ + 9m³- 16m² + 5m

Therefore, standard form of the given polynomial is (-8m⁵ + 10m⁴ + 9m³- 16m² + 5m)

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F(x)=8x^2-2x+3<br> g(x)=12x^2+4x-3<br> what is h(x)=f(x)-g(x)?

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Answer: $h(x)=-4x^2-6x+6$

Step-by-step explanation:

To find the function h(x), you need to subtract the function $f(x)=8x^2-2x+3$ and the function $g(x)=12x^2+4x-3$ .

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If the second term of a geometric sequence of real numbers is -2 and the fifth term is 16 then what is the fourteenth term?

Agata [3.3K]

<h3>Given</h3>

A geometric sequence such that ...

$a_2=-2\quad\text{and}\quad a_5=16$

$a_{14}$

<h3>Solution</h3>

We can use the ratio of the given terms to find the common ratio of the sequence, then use that to find the desired term from one of the given terms. We don't actually need the common ratio (-2). All we need is its cube (-8).

$a_2=a_1r^{(2-1)}=a_1r^1\\a_5=a_1r^{(5-1)}=a_1r^4\\a_{14}=a_1r^{(14-1)}=a_1r^13=a_5r^9\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r^1}=r^3=\dfrac{16}{-2}=-8\\\\r^9=\left(r^3\right)^3=(-8)^3=-512\\a_{14}=a_5(-512)\\\\a_{14}=-8192$

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3 years ago