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3 years ago

12

Can somebody help me find the zeroes for this, algebraically?

1 answer:

Natalija [7]3 years ago

8 0

To solve this algebraically, you can first factor out a 3:
$3(x^2-2x-8)$

Then, you can factor (I'm using guess and check) and my answer is -4 and 2: $3(x-4)(x+2)$

From here on, finding 0s is very easy. You just solve for x in the parentheses. So, the 0s are when x = 4, and x = -2. Remember, (4-4) = 0, and (-2+2) = 0, that's what you want.

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What is <br><img src="https://tex.z-dn.net/?f=8.4%20-%204.2" id="TexFormula1" title="8.4 - 4.2" alt="8.4 - 4.2" align="absmiddle

STALIN [3.7K]

Answer:

4.2

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\space\space&8&.&4\\ -&4&.&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-2=2$

$\frac{\begin{matrix}\space\space&8&.&\textbf{4}\\ -&4&.&\textbf{2}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{2}\end{matrix}}$

$\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}$

$\frac{\begin{matrix}\space\space&8&\textbf{.}&4\\ -&4&\textbf{.}&2\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{.}&2\end{matrix}}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:8-4=4$

$\frac{\begin{matrix}\space\space&\textbf{8}&.&4\\ -&\textbf{4}&.&2\end{matrix}}{\begin{matrix}\space\space&\textbf{4}&.&2\end{matrix}}$

$=4.2$

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2 years ago

Triangle FGH with vertices F(1,8), G(5,7), and H(2,3) in the line y=x.

sergiy2304 [10]

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3 years ago

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How does knowing the double 6+6=12 help solve the near double 6+7=13?

Yakvenalex [24]

Because, if 6 + 6 = 12, then in 6 + 7, 7 can be splitted into 6 + 6 + 1, and so you find it easier...!!

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3 years ago

Graph the lines 2x=-y+7 and 2y=-4x+10. What is the solution?

iris [78.8K]

C) no solution

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3 years ago

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Mia found the area of a polygon The area is 32 square cm.

solniwko [45]

Answer:

Only the isosceles trapezoid has an area of 32 cm².

Step-by-step explanation:

Let's calculate the area of each polygon.

For the two triangles we have:

$A_{t} = \frac{bh}{2} = \frac{2 cm*4 cm}{2} = 4 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle we have:

$A_{r} = bh = 6 cm*4 cm = 24 cm^{2}$

This polygon does not have an area of 32 cm².

For the rectangle trapezoid we have:

$A_{rt} = A_{t} + A_{r} = (4 + 24) cm^{2} = 28 cm^{2}$

So, this polygon does not have an area of 32 cm².

Finally, for the isosceles trapezoid:

$A_{it} = A_{t}*2 + A_{r} = (4*2 + 24) cm^{2} = 32 cm^{2}$

This polygon does have an area of 32 cm².

Therefore, only the isosceles trapezoid has an area of 32 cm².

I hope it helps you!

8 0

3 years ago