Pusa, Delhi, India - 110012.
Verified
Details verified of Geeta M.✕
Identity
Education
Know how UrbanPro verifies Tutor details
Identity is verified based on matching the details uploaded by the Tutor with government databases.
Hindi Mother Tongue (Native)
Delhi University 1977
Bachelor of Arts (B.A.)
Yogrish Astro Research Centre 2023
Advance Vedic Astrology
Pusa, Delhi, India - 110012
ID Verified
Education Verified
Phone Verified
Email Verified
Report this Profile
Is this listing inaccurate or duplicate? Any other problem?
Please tell us about the problem and we will fix it.
Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Astrology Classes
3
1. Which classes do you teach?
I teach Astrology Class.
2. Do you provide a demo class?
Yes, I provide a free demo class.
3. How many years of experience do you have?
I have been teaching for 3 years.
Answered on 26 Feb Learn CBSE/Class 8/Maths/Factorization
To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:
(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq
In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).
Let's find pp and qq:
p+q=6p+q=6 pq=8pq=8
The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.
Therefore, the factorization is:
x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)
Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Astrology Classes
3
Answered on 26 Feb Learn CBSE/Class 8/Maths/Factorization
To factorize the quadratic expression x2+6x+8x2+6x+8, we're looking for two binomials of the form (x+p)(x+q)(x+p)(x+q) where pp and qq are numbers such that:
(x+p)(x+q)=x2+(p+q)x+pq(x+p)(x+q)=x2+(p+q)x+pq
In this case, we want p+qp+q to be equal to the coefficient of xx in the given expression (which is 6) and pqpq to be equal to the constant term (which is 8).
Let's find pp and qq:
p+q=6p+q=6 pq=8pq=8
The pairs of numbers that satisfy these conditions are p=2p=2 and q=4q=4.
Therefore, the factorization is:
x2+6x+8=(x+2)(x+4)x2+6x+8=(x+2)(x+4)
Post your Learning Need
Let us shortlist and give the best tutors and institutes.
or
Send Enquiry to Geeta M.
Let Geeta M. know you are interested in their class
Reply to 's review
Enter your reply*
Your reply has been successfully submitted.