Complete factor the given polynomials. If a polynomial cannot be factored, write "not factorable."

1 answer:

8 0

$6 x^{2} +24x+21$

Multiply 6 and 21 and you get 126. You will need to find two factors of 126 which will add up to 24 (the middle number of the equation). After guess and check, the two numbers are 14 and 9. Instead of 24x, replace it with 14x and 9x.

$6 x^{2} +14x+9x+21$

Factorise each pair separately,

$2x(3x+7) + 3(3x+7)$
$(3x+7)(2x+3)$

Hope this helped!

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Answer:

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Write an inequality that represents eight less than 3 times a number is more than 22

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Answer:

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Step-by-step explanation:

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Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

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The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$