Question

<br />State factor theorem using this theorem, factorise x³-3x²-x+3​

Answer 1

Step-by-step explanation:

first we. will check if x=1 satisfies the above equation.

=(1)^3-3(1)^2-1+3

=1-3-1+3

=0.

hence,x=1 satisfies the eq.

therefore,(x-1) is a factor of given eq.

now we will divide the given eq. by (x-1) to obyain other two factors.

on dividing we will have (x^2-2x-3) as the quotient and 0 as remainder.

now,we know that dividend=(divisor*quotient)+remainder.

x^3-3x^2-x+3={(x-1)(x^2-2x-3)}+0,

now we have to factorise the quotient to get other two factors.

x^2-2x-3=0,

x^2-3x+1x-3=0,

x(x-3)+1(x-3)=0,

(x+1)(x-3)=0.

hence,the other two factors are (x+1)(x-3).

${x}^{3} - 3 {x}^{2} - x + 3 = (x - 1)(x - 3)(x + 1)$

hope this helps.

for division part see the pic.