Question

without actual division prove that 2 x to the power 4 minus 6 x cube + 3 X square + 3 x minus 2 is exactly divisible by x square - 3 x + 2

Answer 1

f(x) = 2x^4 - 3x^3 + 3x - 2 

f(x) = 2*(½)^4 - 3*(½)^3 + 3*(½) - 2 

f(x) = 2*(1/16) - 3*(1/8) + 3*(½) - 2 

f(x) = 2/16 – 3/8 + 3/2 - 2 

f(x) = -3/4 


NO, x = +½ is NOT a root 

Is x = -½ a rational root ? 

f(x) = 2x^4 - 3x^3 + 3x - 2 

f(x) = 2*(-½)^4 - 3*(-½)^3 + 3*(-½) - 2 

f(x) = 2*(1/16) - 3*(-1/8) + 3*(-½) - 2 

f(x) = 2/16 + 3/8 - 3/2 - 2 

f(x) = -3 


NO, x = -½ is NOT a root 

Is x = +2 a rational root ? 

f(x) = 2x^4 - 3x^3 + 3x - 2 

f(x) = 2*2^4 - 3*2^3 + 3*2 - 2 

f(x) = 2*16 - 3*8 + 3*2 - 2 


f(x) = 32 - 24 + 6 - 2 

f(x) = 12 


NO, x = +2 is NOT a root 


Is x = -2 a rational root ? 

f(x) = 2x^4 - 3x^3 + 3x - 2 

f(x) = 2*(-2)^4 - 3*(-2)^3 + 3*(-2) - 2 

f(x) = 2*(16) - 3*(-8) + 3*(-2) - 2 

f(x) = 32 + 24 - 6 - 2 

f(x) = 48 


NO, x = -2 is NOT a root 


Two rational roots have been identified 
rational roots = {+1, -1} 

original polynomial 

2x^4 - 3x^3 + 3x - 2 

Factored 

(x - 1) * (x + 1) * (another factor) 

Is another factor = x^2 - 3x + 2 ? 

(x^2 - 3x + 2 is the factor listed in original problem statement) 

Ordinarily, you would divide to obtain the expression for “another factor”. 

However, the problem states that division is not to be used to solve this problem. 

This does not prevent you from using multiplication. 

If you substitute x^2 - 3x + 2 for “another factor”, and then multiply, you can see if the result is the original polynomial which appears in the problem statement. 

f(x) = (x - 1) * (x + 1) * (x^2 - 3x + 2 ) 

f(x) = (x - 1) * (x + 1) * (x - 2 )* (x - 1) 

Obviously x^2 - 3x + 2 is not a valid factor since its factored form is (x -2 )* (x - 1). 

We have already established that neither x = -2 or x = +2 are not valid rational roots of: 

f(x) = 2x^4 - 6x^3 + 3x^3 + 3x-2 

Therefore, 

(2x^4-6x^3+3x^3+3x-2) is NOT exactly DIVISIBLE by (x^2-3x+2).