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3 years ago

8

Uhhh 6th grade work pls help

2 answers:

Yakvenalex [24]3 years ago

7 0

5b- 1z would be the answer to this question

Anna71 [15]3 years ago

3 0

You just had to distribute it 25b-5z=5(5b-1z)

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Given f(x) = x^3 + kx + 2, and x + 1 is a factor of f(x), then what is the value of <br> k?

Aleks [24]

Answer:

k = 1

Step-by-step explanation:

Given that,

$f(x) = x^3 +kx + 2$

If (x+1) is a factor of (x), we need to find the value of k.

Using remainder theorem to find it.

Put (x+1) = 0

x = -1

Put x = -1 in equation below.

$x^3 + kx + 2=0\\\\(-1)^3+k(-1)+2=0\\\\-1-k+2=0\\\\1-k=0\\\\k=1$

So, the value of k is 1.

4 0

3 years ago

Find all solutions to the equation. -\dfrac{2}{x(x+2)}=\dfrac{x+3}{x+2}− x(x+2) 2 = x+2 x+3 minus, start fraction, 2, divide

Gnoma [55]

Answer:

$x=-1$

Step-by-step explanation:

We have been given an equation $\frac{-2}{x(x+2)}=\frac{x+3}{x+2}$ . We are asked to find the solutions of our given equation.

First of all, we will cross multiply our given equation as:

$-2(x+2)=x(x+2)(x+3)$

Subtract $x(x+2)(x+3)$ from both sides:

$-2(x+2)-x(x+2)(x+3)=x(x+2)(x+3)-x(x+2)(x+3)$

$-2(x+2)-x(x+2)(x+3)=0$

Now we will factor out $(x+2)$ .

$(x+2)(-2-x(x+3))=0$

$(x+2)(-2-x^2-3x)=0$

$(x+2)(-(x^2+3x+2))=0$

Now we will factor by splitting the middle term.

$-(x+2)(x^2+2x+x+2)=0$

$-(x+2)(x+2)(x+1)=0$

$-(x+2)^2(x+1)=0$

Now we will use zero product property and solve for x as:

$-(x+2)^2=0,(x+1)=0$

$(x+2)=0,(x+1)=0$

$x=-2,x=-1$

Now, we will check for extraneous solutions.

We can see that $x=-2$ will make both denominators zero because our function is not defined at $x=-2$ .

Therefore, the solution for our given equation is $x=-1$ .

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4 years ago

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3 years ago

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