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Paul [167]
3 years ago
5

What are the possible values of x in 6x^2 + 432 = 0

Mathematics
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

x = ±6i sqrt(2)

Step-by-step explanation:

6x^2 + 432 = 0

Subtract 432 from each side

6x^2 =- 432

Divide by 6

6x^2 /6 =- 432/6

x^2 =-72

Take the square root of each side

sqrt(x^2) = ±sqrt( -72)

x = ±sqrt(-1) sqrt(2*36)

We know the square root of -1 = i

x = ±i sqrt(2*36)

x = ±i sqrt(36) sqrt(2)

x = ±6i sqrt(2)

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Use the Factor Theorem to determine whether the first polynomial is a factor of the second polynomial.
pashok25 [27]

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Step-by-step explanation:

Factor theorem states that if you divide a polynomial p(x) by a factor x -a of that polynomial, then you will get a zero remainder.

i.,e p(x) = (x-a)q(x)   which means that if x - a is a factor of p(x), then the remainder, when we do synthetic division by  x= a, will be zero.

Determine whether the first polynomial is a factor of the second polynomial.

Given the polynomial:  f(x)=3x^2 + 7x + 40

For  x-5 to be a factor of f(x)=3x^2 + 7x + 40, the factor theorems implies that x = 5 must be a zero of f(x).

Now, to test whether x-5  is a factor;

Set x -5 = 0

⇒x = 5

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we will use synthetic division method to divide f(x) by x =5

you can see the figure as shown below in the attachment.

Since, the remainder is 150 which is not equal to zero, then Factor theorem says that (x-5) is not a  factor of 3x^2 + 7x + 40



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