What is the method for solving the problem of class 9th ncert book of chapter Polynomials chapter of exercise 2.4 of math, give me the best way for solving this question Factorize:(iii) x3+13×2+32x+20
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Let p(x) = x3+13x2+32x+20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x)= x3+13x2+32x+20
p(-1) = (−1)3+13(−1)2+32(−1)+20
= −1+13−32+20
= 0
Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor × Quotient +Remainder
(x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20)
= (x−5)x(x+2)+10(x+2)
= (x−5)(x+2)(x+10)