Q. 1 D5.0( 3 Votes )

Write each polyno

Answer :

Given, p(X) = x2 + 13x + 12


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


x2 + 13x + 12 = 0


the given equation can be written as follows


(x + 12)(x + 1) = 0


Since, we know that x2 + 12x + 13 can be written as (x + a)(x + b)


x2 + 13x + 12 = x2 + (a + b)x + ab


coefficient on either side of the polynomial is same and a + b = 7 and ab = 12


we must find two numbers that satisfy a + b and ab we get, 12 and 1 as the numbers


(x + 12)(x + 1) are the product of first degree polynomial


P(x) = 0 if (x + 12) is 0 and (x + 1) is 0


x + 1 = 0


x = – 1


x2 + 13x + 12 = (– 1)2 + 13(– 1) + 12 = 1 – 13 + 12 = 0


And x + 12 = 0


x = – 12


x2 + 13x + 12 = (– 12)2 + 13(– 12) + 12 = 144 – 156 + 12 = 0


Hence, (x + 12)(x + 1) are the first degree factors of the polynomial and – 12, – 1 are the solutions of the given polynomial x2 + 13x + 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