Question

Factorise X^3-6x^2+3x+10 by using reminder therom

Answer 1

x³-6x²+3x+10
Apply trial and error to the equation by substituting x as any random integer to the equation x3−6x2+3x+10=0x3−6x2+3x+10=0. If the equation gives a result of 0, then (x - root) will be one factor of the equation. 
 
In that case, consider x = 2, so 23−6(2)2+3(2)+10=023−6(2)2+3(2)+10=0 
8−24+6+10=08−24+6+10=0 
 
Hence, (x - 2) is one of the roots of the equation. 
 
Now, apply long division, and accordingly we will obtain another factor as x2−4x−5=(x−5)(x+1)x2−4x−5=(x−5)(x+1). 
 
Hence, x3−6x2+3x+10=(x+1)(x−2)(x−5)x3−6x2+3x+10=(x+1)(x−2)(x−5).