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3 years ago

Factor -2bk^2 + 6bk - 2b. -2b(k^2 - 3k - 1) -2b(k^2 + 3k + 1) -2b(k^2 - 3k + 1)

Mathematics

2 answers:

Aleks [24]3 years ago

7 0

Answer:

-2b(k²-3k+1)

Step-by-step explanation:

The question is on factorization

-2bk²+6bk-2b

Factor out 2b

-2b(k²-3k+1)

When you open brackets to check the answer

-2b×k²-2b×-3k-2b×1= -2bk²+6bk+2b

Nikitich [7]3 years ago

5 0

For this case we must factor the following expression:

$-2bk ^ 2 + 6bk-2b$

It is observed that the three terms have in common a ce divisible between 2 and the variable "b".

Then, we can extract common factor "2b":

$2b (-k ^ 2 + 3k-1)$

If we take out common factor "-" we have:

$-2b (k ^ 2-3k + 1)$

Answer:

Option C

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1. Find the vertex of the following: y = (x - 4)(x - 10) *

horsena [70]

Answer:

1. (7,-9)

2. (4,0) and (10,0)

3. (3,0) and (-3,0)

4. (0,-9)

5. (-1,-32)

Step-by-step explanation:

To find the vertex use the formula

$\frac{-b}{2a}$

First make the equation into a quadratic equation.

$(x-4)(x-10)\\x^{2} -10x-4x+40\\x^{2} -14x+40$

Now it is in the form

$ax^{2} +bx+c=0$

Now we can substitute the values

$\frac{-(-14)}{2(1)} \\\frac{14}{2} \\7$

Now we have 7 as our x value for the vertex we can substitute for y,

$x^{2} -14x+40=y\\(7)^{2} -14(7)+40=y\\49-98+4=y\\y=-9$

So the vertex for 1 is (7,-9)

To find the intercepts for 2 & 3, they basically already give it to you. You just need to find a value for x for x-4 that will equal it to 0 and another for x for x-10 that will equal it to 0.

$x-4=0\\x=4\\\\x-10=0\\x=10$

4 is slightly different but start the same,

$(x-3)(x+3)\\x^{2}+3x-3x-9\\x^{2}-9$

Here, c (9) just shows the graph move down 9, so the y intercept = 9.

5 is the same process as 1,

$2(x-3)(x+5)\\2(x^{2}+5x-3x-15)\\2(x^{2}+2x-15)\\2x^{2}+4x-30$

Vertex,

$\frac{-(4)}{2(2)} \\\frac{-4}{4} \\-1$

Substitute,

$2x^{2}+4x-30\\2(-1)^{2}+4(-1)-30\\2-4-30\\-32$

So the vertex is (-1,-32)

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