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Kamila [148]
3 years ago
7

Write an equation in standard form with zeros 2 and -8

Mathematics
1 answer:
Komok [63]3 years ago
8 0

Answer:

Standard form  of polynomial with zeros 2 and -8 isP(x) = x^2 + 6x - 16

Step-by-step explanation:

Here, the zeroes of the polynomial P(x) is given as  x = 2 and x  = -3

If x  = a is a zero pf the polynomial p(x) , then (x-a) is the root of the given polynomial.

⇒ (x- 2)  and (x -(-8)) are the two roots of the polynomial P(x).

Now, Polynomial P(x) = Product of all its Roots

⇒P(x)  =  (x- 2) (x +8 )\\= x(x+8) -2(x+8)  = x^2 + 8x - 2x- 16\\= x^2 + 6x - 16 \\\implies P(x) = x^2 + 6x - 16

Hence the standard form  of polynomial with zeros 2 and -8 isP(x) = x^2 + 6x - 16

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For each function, find f(-1), f(1), and f(3)<br> f(x)=3x-4
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Answer:

f(x)= 3x-4

f(-1)= 3*-1-4

f(-1)= 2-4

f(-1)= -2

f(x)= 3x-4

f(3)= 3*3-4

f(3)= 9-4

f(3)=5

f(x)= 3x-4

f(1)= 3*1-4

f(1)= 3-4

f(1)= -1

Step-by-step explanation:

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Obtain the general solution:<br><br> (x-y)(4x+y)dx + x(5x-y)dy = 0
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Please help asap <br> (possible answers are: 2,-5,1,-8)
mojhsa [17]

The solution is y = -5

What is Quadratic Equation?

A quadratic equation is a second-order polynomial equation in a single variable x , ax² + bx + c = 0. with a ≠ 0. Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has at least one solution. The solution may be real or complex.

Given data ,

Let the function be f ( x ) = y

And , y = x² + 2x - 8

So , f ( x ) = x² + 2x - 8

Substituting the value for x in the equation , we get

When x = -3

f ( -3 ) = ( -3 )² + 2( -3 ) - 8

         = 9 - 6 - 8

         = 9 - 14

f ( -3 ) = -5

When x = 1

f ( 1 ) = ( 1 )² + 2( 1 ) - 8

       = 1 + 2 - 8

       = 3 - 8

f ( 1 ) = -5

Hence , the value of y = -5

To learn more about quadratic equation click :

brainly.com/question/25652857

#SPJ1

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