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7 months ago

Factor to the given polynomial completely. 4x³+24x²-288x

Mathematics

2 answers:

STatiana [176]7 months ago

3 0

Answer: $\Large\boxed{4x(x+12)(x-6)}$

Step-by-step explanation:

Given expression

4x³ + 24x² - 288x

Factorize 4x out from the expression

4x · x² + 4x · 6x - 4x · 72

4x (x² + 6x - 72)

Cross multiply to factorize the remaining polynomial expression

The meaning is to allow the factored product of the constant to add up to the first-degree term

x 12

x -6

Combine the result

$\Large\boxed{4x(x+12)(x-6)}$

Hope this helps!! :)

Please let me know if you have any questions

arlik [135]7 months ago

3 0

Answer:

4x(x + 6)(x - 12).

Step-by-step explanation:

First take out 4x which is the GCF of the 3 terms:

= 4x (x^2 - 6x - 72)

We need 2 numbers whose product is -72 and whose sum is -6.

They are -12 and + 6 so we have:

4x(x + 6)(x - 12)

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2 years ago

Alex ate 4 less the 2 times the slices of pizza that Oliver ate.If Oliver ate 8 slices of pizza how many slices did alex eat?

Fofino [41]

Answer: 12

Step-by-step explanation: 2 times of 8 is 16, but it says 4 less, so you subtract 4 and that equals 12.

7 0

1 year ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

Answer-

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

Solution-

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

The distance between bases on a baseball field is 27.43 meters.

Mkey [24]

27.43 * 4.5 = 123.435

Joe jogged 123.435 meters

8 0

2 years ago

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Harlamova29_29 [7]

4440/10 Is the answer

6 0

2 years ago

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