Q. 7 F4.3( 4 Votes )

# Find the zeroes o

For finding zeroes, we’ll use the hit and trial method.

For that we’ll put a value which will make our polynomial zero.

Since, here the whole equation is combination of positive and negative that means, zero can be achieved by putting the positive or negative values.

So, first we’ll put x = 1.

(1)3 – 23(1)2 + 142(1) – 120 = 1 – 23 + 142 – 120

= 0

x = 1 is one of the zeroes.

Now, we’ll write the equation in terms of (x – 1) to make equation quadratic and hence to find the other zeroes.

x3 – 23x2 + 142x – 120 = x2 (x – 1) – 22x(x – 1) + 120(x – 1)

= (x – 1)(x2 – 22x + 120)

Now, using factorization by splitting the middle term method -

= (x – 1)(x2 – (12 + 10)x + 120)

= (x – 1)(x2 – 12x – 10x + 120)

= (x – 1)(x(x – 12) – 10(x – 12))

= (x – 1)(x – 10)(x – 12)

Now, to get the zeroes we’ll put these equals to zero.

x – 1 = 0, x – 10 = 0, x – 12 = 0

x = 1, x = 10, x = 12 are the zeroes of x3 – 23x2 + 142x – 120

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 