NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3 Solutions are based on the concept of Remainder Theorem.

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Check out NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 Solutions

Read More: NCERT Solutions For Class 9 Maths Chapter 2 Polynomials

Exercise Solutions of Class 9 Maths Chapter 2 Polynomials

Also check other Exercise Solutions of Class 9 Maths Chapter 2 Polynomials

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Chapter Related Topics
Polynomials	Remainder Theorem	Degree of polynomial
Polynomial Important Questions	Polynomials in One Variable	Synthetic Division
Remainder	Algebraic Identities	Cubic Polynomials

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CBSE X Related Questions

1.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
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2.
In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :
- \( 130^\circ \)
- \( 50^\circ \)
- \( 120^\circ \)
- \( 90^\circ \)
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3.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
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4.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
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5.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
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6.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution

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