actor x3 + x2 + x + 1 by grouping. What is the resulting expression? (x2)(x + 1) (x2 + 1)(x) (x2 + 1)(x + 1) (x3 + 1)(x + 1)

1 answer:

5 0

So for this, factor x^3 + x^2 and x + 1 separately. Make sure that they have the same quantity on the inside: $x^2(x+1)+1(x+1)$

Now you can rewrite the expression as: $(x^2+1)(x+1)$ , which is your final answer.

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13. Danny is in science class and is given a square block of metal with

jenyasd209 [6]

Answer: 336g.

Step-by-step explanation:

the formula for density is mass divided by volume (m/v). In this problem, you mus first find the volume of the square block. The formula for volume is length multiplied by width multiplied by height (LxWxH). To solve for volume you plug in the values to the equation: 6x4x2 = 48.

Now, you have the volume, and you have the density. Now you need to solve for mass. Plug in the values you have into the density equation (d=m/v):

7=m/48

to isolate M, multiply each side by 48. this leaves you with 7x48 = m. now all thats left is to solve. 7x48 = 336g. You can double check by plugging the values back into the equation. 336/48 = 7.

5 0

3 years ago

Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Blizzard [7]

Answer:

One possible equation is $f(x) = (x + 3)\, (x - 5)$ , which is equivalent to $f(x) = x^{2} - 2\, x - 15$ .

Step-by-step explanation:

The factor theorem states that if $x = x_{0}$ (where $x_{0}$ is a constant) is a root of a function, $(x - x_{0})$ would be a factor of that function.

The question states that $(-3,\, 0)$ and $(5,\, 0)$ are $x$ -intercepts of this function. In other words, $x = -3$ and $x = 5$ would both set the value of this quadratic function to $0$ . Thus, $x = -3\!$ and $x = 5\!$ would be two roots of this function.

By the factor theorem, $(x - (-3))$ and $(x - 5)$ would be two factors of this function.

Because the function in this question is quadratic, $(x - (-3))$ and $(x - 5)$ would be the only two factors of this function. In other words, for some constant $a$ ( $a \ne 0$ ):

$f(x) = a\, (x - (-3))\, (x - 5)$ .

Simplify to obtain:

$f(x) = a\, (x + 3)\, (x - 5)$ .

Expand this expression to obtain:

$f(x) = a\, (x^{2} - 2\, x - 15)$ .

(Quadratic functions are polynomials of degree two. If this function has any factor other than $(x - (-3))$ and $(x - 5)$ , expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of $a$ corresponds to a distinct quadratic function with $x$ -intercepts $(-3,\, 0)$ and $(5,\, 0)$ . For example, with $a = 1$ :