Answer :

Given, p(X) = x2 – 8x + 12


Now, we need to write the given polynomial as a product of first degree polynomial and p(x) = 0


x2 – 8x + 12 = 0


The given equation can be written as follows


(x – 6)(x – 2) = 0


Since, we know that x2 – 8x + 12 can be written as (x – a)(x – b)


x2 – 8x + 12 = x2 – (a + b)x + ab


coefficient on either side of the polynomial is same and a + b = 8 and ab = 12


we must find two numbers that satisfy a + b and ab we get, 6 and 2 as the numbers


(x – 6)(x – 2) are the product of first degree polynomial


P(x) = 0 if (x – 6) is 0 and (x – 2) is 0


x – 6 = 0


x = 6


x2 – 8x + 12 = 62 – 8(6) + 12 = 36 – 48 + 12 = 0


And x – 2 = 0


x = 2


x2 – 8x + 12 = 22 – 8(2) + 12 = 4 – 16 + 12 = 0


Hence, (x – 6)(x – 2) are the first degree factors of the polynomial and 6, 2 are the solutions of the given polynomial x2 – 8x + 12


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
caricature
view all courses
RELATED QUESTIONS :

A square ground hKerala Board Mathematics Part-1

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

Write each polynoKerala Board Mathematics Part-2

By adding a numbeKerala Board Mathematics Part-2

By adding a numbeKerala Board Mathematics Part-2

Find a third degrKerala Board Mathematics Part-2

Find a second degKerala Board Mathematics Part-2