Question

If 1 and 3 are zeros of polynomial p(x) then find the remaining zeros of p(x)
p(x)=2x^4-7x^3-13x^2+63x-45

Answer 1

Answer:

5/2 and -3

( If you need it written differently, notice that 5/2 = 2.5 )

Step-by-step explanation:

Let the remaining zeros be a and b.

From the coefficients of x³ and x⁴, the sum of all four roots is 7/2.  So...

  a + b + 1 + 3 = 7/2  ⇒  a + b = -1/2.

From the  coefficient of x⁴ and the constant coefficient, the product of the roots is -45/2.  So...

 3ab = -45/2  ⇒  ab = -15/2.

It follows that a and b are the roots of the quadratic

      2x² + x - 15 = 0

⇒  ( 2x - 5 ) ( x + 3 ) = 0

⇒  x = 5/2  or  x = -3.

So the two roots a and b are 5/2 and -3.

Hope this helps.

Answer 2

with explained answer

for u

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