10 points high school polynomial question?

Question

given P(x) = 2x^3 - 7x^2 + 4x + 1

if the equation P (x) = 0 has a root at x = 1 , find the sum and product of the other roots

Answer 1

Since x=1 satisfies p(x) = 0 ; (x -1) is a factor of p(x)

divide p(x) by x-1 to get

p(x) = (x-1)(2x^2-5x-1)

Putting p(x)=0

(x-1)(2x^2 - 5x - 1) = 0

x = 1 or 2x^2 - 5x - 1 = 0

Sum and product of roots of 2x^2 - 5x - 1 are 2.5 and (-0.5) respectively.

Roots of p(x) are    1     AND      {5+(33^1/2)} / 4         AND

{5 - (33^1/2)} / 4

Answer 2

P(x) = 2x^3 - 7x^2 + 4x + 1

One root is x=1

Therefore (x-1) is a factor

P(x)=(x-1) (2x^2 - 5x - 1)

The above step can be completed through inspection, by looking at the coefficients. However, long division can also be used (although it is a more tedious method).

Since;

P(x)

=(x-1) (2x^2 - 5x - 1)

=0.5 (x-1) (x^2 - 5/2x - 1/2)

=0.5 (x-1) [(x- 5/4)^2 - 25/16 - 1/2]

=0.5 (x-1) [(x- 5/4)^2 - 33/16]

=0.5 (x-1) [(x- 5/4) + sqrt(33)/4] [(x- 5/4) - sqrt(33)/4]

The other roots could also have been found by applying the quadratic formula to (2x^2 - 5x - 1)

Therefore the roots are 1, 5/4 - sqrt(33)/4, 5/4 + sqrt(33)/4

Which can be expressed as 1, (5 - sqrt(33))/4, (5 + sqrt(33))/4

The sum of the two other roots we have just found is;

(5 - sqrt(33))/4 + (5 + sqrt(33))/4

=10/4

=5/2

The product of the two other roots we have just found is;

(5 - sqrt(33))/4 * (5 + sqrt(33))/4

=(25 - 33) / 4 -------> difference of perfect squares

= -8/16

= -1/2