What is the remainder when (3x^4+2x^3-x^2+2x-19)÷(x+2)

Mathematics

1 answer:

Solnce55 [7]3 years ago

3 0

I believe the remainder would be 5.

Hope this helps!

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What are the ordered pairs of the

9966 [12]

Start by writing the system down, I will use $y$ to represent $f(x)$

$y=x^2-5x+3\wedge y=-3$

Substitute the fact that $y=-3$ into the first equation to get,

$-3=x^2-5x+3$

Simplify into a quadratic form ( $ax^2+bx+c=0$ ),

$x^2-5x+6=0$

Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,

$x^2+x(m+n)+mn=0$

which then must factor into

$(x+m)(x+n)=0$

And the solutions will be $m,n$ .

Clearly for small coefficients like ours $5,6$ , this is very easy to figure out. To get 5 and 6 we simply say that $m=3, n=2$ .

This fits the definition as $5=3+2$ and $6=2\cdot3$ .

So as mentioned, solutions will equal to $m=3, n=2$ but these are just x-values in the solution pairs of a form $(x,y)$ .

To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.

So the solution pairs are $(2,0)$ and $(3,0)$ .

Hope this helps :)

5 0

3 years ago

Graph the solution of 2 ≥ 4 - v

Lerok [7]

For this case we have the following inequality:
2 ≥ 4 - v
The first thing we must do in this case is to clear the value of v.
We have then:
v ≥ 4 - 2
v ≥ 2
Therefore, the solution set is given by:
[2, inf)
Answer:
See attached image.