Question

please solve this question

Answer 1

Hello, 

q(x) = x³ + 2x² + a 
p(x) = x⁵ - x⁴ - 4x³ + 3x² + 3x + b 

p(x) = x⁵ - x⁴ - 4x³ + 3x² + 3x + b 
   = x⁵ + 2x⁴ + ax² - 3x⁴ - 4x³ + (3 - a)x² + 3x + b 
   = x²(x³ + 2x² + a) - 3x⁴ - 6x³ - 3ax + 2x³ + (3 - a)x² + 3(1 + a)x + b 
   = x²(x³ + 2x² + a) - 3x(x³ + 2x² + a) + 2x³ + 4x² + 2a - (a + 1)x² + 3(1 + a)x + (b - 2a) 
   = x²(x³ + 2x² + a) - 3x(x³ + 2x² + a) + 2((x³ + 2x² + a) - (a + 1)x² + 3(1 + a)x + (b - 2a) 
   = (x³ + 2x² + a)(x² - 3x + 2) - (a + 1)x² + 3(1 + a)x + (b - 2a) 
   = (x² - 3x + 2).q(x) - (a + 1)x² + 3(1 + a)x + (b - 2a) 

If p(x) and q(x) share the same roots, then p(x) is divisible by q(x). 
If p(x) is divisible by q(x), the remainder -(a+1)x²+3(1+a)x+(b-2a) must be nil for any x. 

This means each and every coefficient of this remainder polynomial must be nil: 
{ -(a + 1) = 0 
{ 3(1 + a) = 0 
{ b - 2a = 0 

The solution is obviously: 
{ a = -1 
{ b = -2 

With this values of a and b, we would get: 
p(x) = (x² - 3x + 2).q(x) 
and thus would the roots of q(x) forcefully be also roots of p(x). 

And the roots of p(x) that are not roots of q(x) are evidently the roots of: 
x² - 3x + 2 = 0 
x² - 2x - x + 2 = 0 
x(x - 2) - (x - 2) = 0 
(x - 2)(x - 1) = 0 
And they are x=1 or x=2 

q(x) = x³ + 2x² - 1 
   = x³ + x² + x² - 1 
   = x²(x + 1) + (x - 1)(x + 1) 
   = (x + 1)(x² + x - 1) 
   = (x + 1)[x + 2x(½) + (½)² - 5/4] 
   = (x + 1)[(x + ½)² - ((√5)/2)²] 
   = (x + 1)[x + (1 - √5)/2)][x + (1 + √5)/2)] 

Conclusion: 
a=-1 and b=-2 

Roots of q(x): { -1; (-1 + √5)/2; (-1 - √5)/2 } 
Roots of p(x): { 1; 2; -1; (-1 + √5)/2; (-1 - √5)/2 }