Math, asked by chintalaRevathi, 4 months ago

EXERCISE - 2.4
1. Determine which of the following polynomials has (x + 1) as a factor.
3--x+1
(i) x4 - x + x2 - x + 1
(i) x + 2x + 2x2 + x + 1
(iv) x - x2 - (3 - 13)x+ V3
2. Use the Factor Theorem to determine whether g(x) is factor of f(x) in each of
following cases
(6) f(x) = 5x + x? - 5x - 1, g(x) = x + 1
(ii
) f(x) = x + 3x2 + 3x + 1, g(x) = x + 1
(ii) f(x) = x - 4x² + x + 6, g(x) = x - 2
(iv) f(x) = 3x + x2 - 20x + 12, g(x) = 3x - 2
(1) Ax) = 4x + 20x² + 33x + 18, g(x) = 2x + 3
3. Show that (x - 2), (x + 3) and (x - 4) are factors of x - 3x2 - 10x + 24.
Show that (x + 4), (x - 3) and (x - 7) are factors of x2 - 6x² - 19x + 84.
Gif both (x - 2) and
are factors of px? + 5x + r, show that p=r.
2​

Answers

Answered by Ritikabaghav23
3

Answer:

1)Apply remainder theorem

x+1=0

x=−1

Put the value of x=−1 in all equations.

(i) x

3

+x

2

+x+1=(−1)

3

+(−1)

2

+(−1)+1=−1+1−1+1=0

Then x+1 is the factor of equation

(ii) x

4

+x

3

+x

2

+x+1=(−1)

4

+(−1)

3

+(−1)

2

+(−1)+1=1−1+1−1+1=1

This is not zero.Then x+1 is not the factor of equation

(iii) x

4

+3x

3

+3x

2

+x+1=(−1)

4

+3(−1)

3

+3(−1)

2

+(−1)+1=1

This is not zero.Then x+1 is not the factor of equation

(iv)x

3

−x

2

−(2+

2

)x+

2

=(−1)

3

−(−1)

2

−(2+

2

)(−1)+

2

=−1−1+2−

2

+

2

=0

This is zero. Then x+1 is the factor of equation

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