Respuesta :

vlatkostojanovic

Answer:

Therefore, the  first polynomial is:

[tex]Z=\frac{1}{2}x^2+\frac{1}{2}\\[/tex]

the second polynomial is:

[tex]Y=\frac{1}{2}x^2-\frac{1}{2}\\[/tex]

Step-by-step explanation:

We assume that: Z is first polynomial and Y is second polynomial.

Therefore, we have

[tex]Z+Y=x^2\\\\Z-Y=1\\\\\implies Z+Y+Z-Y=x^2+1\\\\2Z=x^2+1\\\\Z=\frac{1}{2}x^2+\frac{1}{2}\\[/tex]

now, we get

[tex]\implies Z+Y-Z+Y=x^2-1\\\\2Y=x^2-1\\\\Y=\frac{1}{2}x^2 -\frac{1}{2}[/tex]

Therefore, the  first polynomial is:

[tex]Z=\frac{1}{2}x^2+\frac{1}{2}\\[/tex]

the second polynomial is:

[tex]Y=\frac{1}{2}x^2-\frac{1}{2}\\[/tex]

facundo3141592

We will see that the polynomials are:

  • p(x) = (1/2)*x^2 + 1/2
  • q(x) = (1/2)*x^2 - 1/2

How to find the polynomials?

We want to get two polynomials p(x) and q(x) such that:

  • p(x) + q(x) = x^2
  • p(x) - q(x) = 1

Because of the things that appear on the sum and difference we can assume that:

  • p(x) = a*x^2 + b
  • q(x) = c*x^2 + d

Then when we rewrite the equations above we find that:

p(x) + q(x) = (a + c)*x^2 + (b + d) = x^2

From this we get:

a + c = 1

b + d = 0

The other equation is:

p(x) - q(x) = (a - c)*x^2 + (b - d) = 1

Then we have:

a - c = 0

b - d = 1

Now we just need to find solutions of:

  • a + c = 1
  • b + d = 0
  • a - c = 0
  • b - d = 1

From the second equation we can get:

b = -d

Then we replace that in the fourth equation:

b - d = 1

b + b = 1

2*b = 1

b = 1/2

Then:

d = -b = -1/2

Now two equations remain:

a + c = 1

a - c = 0

From the second we get:

a = c

Replacing that in the first one:

a + c = 1

a + a = 1

2a = 1

a = 1/2 = c

Then the polynomials are:

  • p(x) = (1/2)*x^2 + 1/2
  • q(x) = (1/2)*x^2 - 1/2

If you want to learn more about polynomials, you can read:

https://brainly.com/question/4142886

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