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2 years ago

Factor using the x method ( please do not answer without showing work )

Mathematics

2 answers:

muminat2 years ago

8 0

Answer:

$5(x + 10)(10x - 3)$

Step-by-step explanation:

We are factoring

$50x^{2} + 485x - 150$

So:

((2•5^2x^2) + 485x) - 150

Pull like factors :

50x^2 + 485x - 150 = 5 • (10x^2 + 97x - 30)

Factor

10x^2 + 97x - 30

Step-1: Multiply the coefficient of the first term by the constant 10 • -30 = -300

Step-2: Find two factors of -300 whose sum equals the coefficient of the middle term, which is 97.

-300 + 1 = -299

-150 + 2 = -148

-100 + 3 = -97

-75 + 4 = -71

-60 + 5 = -55

-50 + 6 = -44

-30 + 10 = -20

-25 + 12 = -13

-20 + 15 = -5

-15 + 20 = 5

-12 + 25 = 13

-10 + 30 = 20

-6 + 50 = 44

-5 + 60 = 55

-4 + 75 = 71

-3 + 100 = 97

Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 100

10x^2 - 3x + 100x - 30

Step-4: Add up the first 2 terms, pulling out like factors:

x • (10x-3)

Add up the last 2 terms, pulling out common factors:

10 • (10x-3)

Step-5: Add up the four terms of step 4:

(x+10) • (10x-3)

Which is the desired factorization

Thus your answer is

$5(x + 10)(10x - 3)$

likoan [24]2 years ago

3 0

Answer:

$\displaystyle \rm 5({x}^{} + 10)( 10x - 3)$

Step-by-step explanation:

we would like to factor out the following expression:

$\displaystyle {50x}^{2} + 485x - 150$

notice that, in every term there's a common factor of 5 thus factor it out:

$\displaystyle 5( {10x}^{2} + 97x - 30)$

now we have to rewrite the middle term as sum or substraction of two different terms in that case 100x-3x can be considered:

$\displaystyle 5( {10x}^{2} + 100 - 3x - 30)$

factor out 10x:

$\displaystyle 5( 10({x}^{2} + 10)- 3x - 30)$

factor out -3:

$\displaystyle \rm 5( 10x({x}^{} + 10)- 3(x + 10))$

group:

$\displaystyle \rm 5({x}^{} + 10)( 10x - 3)$

and we are done!

hence,

our answer is B)

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Step-by-step explanation:

See comment for complete question

The question is an illustration of binomial probability where

$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$(a):\ P(x \ge 5)$

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$p = 75\%$ --- proportion that are extroverts

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$P(x \ge 5) = 1 - P(x < 5)$

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$p = 75\%$ --- proportion that are extroverts

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$(c)\ P(x = 0)$

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$p = 70\%$ --- proportion that are introverts

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$P(x) = ^nC_x * p^x * (1 - p)^{n-x}$

$P(x = 0) = ^{5}C_0 * (70\%)^0 * (1 - 70\%)^{5-0}$

$P(x = 0) = 1 * 1 * (0.30)^5$

$P(x = 0) = 0.002$

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