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4 years ago

Factor x3-2x2-9x+18 completely if(x+3) is a factor

Mathematics

2 answers:

Charra [1.4K]4 years ago

6 0

Factor x^3 -2x^2-9x+18

= x^2(x - 2) - 9(x - 2)

= (x^2 - 9)(x -2)

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

So (x+3) is a factor

Phoenix [80]4 years ago

3 0

Dividing:-

x + 3) x^3 - 2x^2 - 9x + 18 { x^2 - 5x + 6

x^3 + 3x^2

----------------

-5x^2 - 9x

-5x^2 - 15x

----------------

6x + 18

.............

The quotient is x^2 - 5x + 6

which is (x - 3)(x - 2)

So the factors of the original expression

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

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Identify the vertex (the 2 beside the "x" means squared) <br><br> y=-x2 + 18x - 72

myrzilka [38]

Answer:

Write the problem as a mathematical expression.

y=−x²+18x−72

Rewrite the equation in vertex form.

y=−(x−9)²+9

Use the vertex form, y=a(x−h)2+k , to determine the values of a, h, and k.

a=−1

h=9

k=9

Find the vertex (h,k).

(9,9)

Good Luck !

Step-by-step explanation:

6 0

3 years ago

And the 95 fifth and sixth graders going on a field trip there are 27 more fifth-graders than six graders how many fifth-graders

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F+s=95
f=s+27
s+27+s=95
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6 0

4 years ago

Read 2 more answers

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Oksana_A [137]

$\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}$

The length of the curve is given by the integral

$\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt$

Expand and rewrite the integrand:

$(e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}$
$=e^{-2t}(e^{4t}+2e^{2t}+1)$
$=e^{-2t}(e^{2t}+1)^2$
$\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}$

Now the integral is

$\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt$
$=2\displaystyle\int_0^3\cosh t\,\mathrm dt$
$=2\sinh t\bigg|_{t=0}^{t=3}$
$=2\sinh 3$