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2 years ago

Which polynomial is prime? Ox+3x²-x²-3 Ox4-3x²-x² + 3 O 3x²+x-6x-2 O 3x²+x-6x + 3

Mathematics

1 answer:

GarryVolchara [31]2 years ago

8 0

Answer:

The polynomial 3x² + x - 6x + 3 is a prime polynomial

How to determine the prime polynomial?

For a polynomial to be prime, it means that the polynomial cannot be divided into factors

From the list of options, the polynomial (D) is prime, and the proof is as follows:

We have:

3x² + x - 6x + 3

From the graph of the polynomial (see attachment), we can see that the function does not cross the x-axis.

Hence, the polynomial 3x² + x - 6x + 3 is a prime polynomial

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What pair of numbers have a sum of 100?

serg [7]

Answer:

50+50

75+25

60+40

90+10

45+55

Step-by-step explanation:

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3 years ago

Help please 50 points!

uranmaximum [27]

Answer:

A and B

Step-by-step explanation:

we would like to solve the following equation:

$\rm\displaystyle 5 - 2x = \sqrt{ {2x}^{2} + x - 1 } - x$

to do so isolate -x to the left hand side and change its sign:

$\rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^{2} + x - 1 }$

simplify addition:

$\rm\displaystyle 5 - x = \sqrt{ {2x}^{2} + x - 1 }$

square both sides:

$\rm\displaystyle (5 - x {)}^{2} = (\sqrt{ {2x}^{2} + x - 1 } {)}^{2}$

simplify square of the right hand side:

$\rm\displaystyle (5 - x {)}^{2} = {2x}^{2} + x - 1$

use (a-b)²=a²-2ab+b² to expand the left hand side:

$\rm\displaystyle {x}^{2} - 10x + 25= {2x}^{2} + x - 1$

swap the equation:

$\rm\displaystyle {2x}^{2} + x - 1 = {x}^{2} - 10x + 25$

isolate the right hand side expression to the left hand side and change every sign:

$\rm\displaystyle {2x}^{2} + x - 1 - {x}^{2} + 10x - 25 = 0$

simplify:

$\rm\displaystyle {x}^{2} + 11x - 26= 0$

rewrite the middle term as 13x-2x:

$\rm\displaystyle {x}^{2} + 13x - 2x - 26= 0$

factor out x:

$\rm\displaystyle x({x}^{} + 13)- 2x - 26= 0$

factor out -2:

$\rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0$

group:

$\rm\displaystyle (x- 2)(x + 13)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}$

solve for x:

$\displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}$

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

$\rm\displaystyle 5 - x \geq0$

solve the inequality for x:

$\rm\displaystyle x \leqslant 5$

since 2 and -13 is less than 5 both solutions are valid for x hence,

$\displaystyle \begin{cases}x _{1} = 2\\ x_{2} = - 13 \end{cases}$

and we're done!

4 0

3 years ago

Read 2 more answers

Point M is the midpoint of segment JK. Find JK when JM=6x-7 and MK=2x+3.

Whitepunk [10]

If M is the midpoint of JK then JM+MK=JK and JM=MK
so
6x-7=2x+3
minus 2x both sides
4x-7=3
add 7 both sides
4x=10
divide both sides by 4
$x=\frac{10}{4}$
$x=\frac{5}{2}$

sub back

JM=6x-7
$JM=6(\frac{5}{2})-7$
$JM=\frac{30}{2}-7$
$JM=15-7$
JM=8

JM=MK=8
JM+MK=8+8=16=JK

$x=\frac{5}{2}$ and JM=8=MK and JK=16

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4 years ago

Read 2 more answers

Me ayudan alguien/ can someone help me!!!

Tresset [83]

Answer:

divide perimeter by 3

Step-by-step explanation:

all 3 sides of an equilateral triange are equal, so you can divide the perimeter by 3 to get the length of one side

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