Question

solve 2y³+y²-2?????​

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Exercise - 2.4

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Polynomials

Question-1 :- Determine which of the following polynomials has (x + 1) a factor :

(i) x³ + x² + x + 1,    (ii) x⁴ + x³ + x² + x + 1,    (iii) x⁴ + 3x³ + 3x² + x + 1,    (iv) x³ - x² - (2 + √2)x + √2.

Solution :-

(i) x³ + x² + x + 1

Let x + 1 is a factor of given polynomial.

Now, x + 1 = 0

x = -1

p(-1) = (-1)³ + (-1)² + (-1) + 1

= -1 + 1 - 1 + 1

= 2 - 2 = 0

Therefore, It is confirmed that x + 1 is a factor of x³ + x² + x + 1.

(ii) x⁴ + x³ + x² + x + 1

Let x + 1 is a factor of given polynomial.

Now, x + 1 = 0

x = -1

p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1

= 1 - 1 + 1 - 1 + 1

= 3 - 2 = 1

Therefore, x + 1 is not a factor of x⁴ + x³ + x² + x + 1.

(iii) x⁴ + 3x³ + 3x² + x + 1

Let x + 1 is a factor of given polynomial.

Now, x + 1 = 0

x = -1

p(-1) = (-1)⁴ + 3 x (-1)³ + 3 x (-1)² + (-1) + 1

= 1 - 3 + 3 - 1 + 1

= 5 - 4 = 1

Therefore, x + 1 is not a factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) x³ - x² - (2 + √2)x + √2

Let x + 1 is a factor of given polynomial.

Now, x + 1 = 0

x = -1

p(-1) = (-1)³ - (-1)² - (2 + √2) x (-1) + √2

= -1 - 1 + 2 + √2 + √2

= -2 + 2 + 2√2

= 2√2

Therefore, x + 1 is not a factor of x³ - x² - (2 + √2)x + √2.

Question-2 :- Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1,

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2,

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3 .

Solution :-

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1,

By factor theorem (x + 1) is a factor of 2x³ + x² – 2x – 1.

If x + 1 = 0

x = -1

p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1

= -2 + 1 + 2 - 1

= 3 - 3 = 0

Therefore, g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1.

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

By factor theorem (x + 2) is a factor of x³ + 3x² + 3x + 1.

If x + 2 = 0

x = -2

p(-1) = 2 x (-2)³ + 3 x (-2)² + 3 x (-2) + 1

= 2 x (-8) + 3 x 4 - 6 + 1

= -16 + 12 - 6 + 1

= -22 + 13 = -9

Therefore, g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1.

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

By factor theorem (x – 3) is a factor of x³ – 4x² + x + 6.

If x – 3 = 0

x = 3

p(3) = (3)³ – 4 x (3)² + 3 + 6

= 27 - 36 + 9

= 36 - 36 = 0

Therefore, g(x) = x – 3 is a factor of p(x) = x³ – 4x² + x + 6.

Question-3 :- Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x² + x + k,    (ii) p(x) = 2x² + kx + √2,    (iii) p(x) = kx² – √2x + 1,    (iv) p(x) = kx² – 3x + k.

Solution :-

(i) p(x) = x² + x + k , k = ?

Given that x - 1 is a factor of p(x).

So, x - 1 = 0

x = 1

And p(1) = 0.

p(1) = (1)² + 1 + k

0 = 1 + 1 + k

0 = 2 + k

k = -2

(ii) p(x) = 2x² + kx + √2, k = ?

Given that x - 1 is a factor of p(x).

So, x - 1 = 0

x = 1

And p(1) = 0.

p(1) = 2 x (1)² + k x 1 + √2

0 = 2 + k + √2

0 = 2 + √2 + k

k = -2 - √2

k = -(2 + √2)

(iii) p(x) = kx² – √2x + 1, k = ?

Given that x - 1 is a factor of p(x).

So, x - 1 = 0

x = 1

And p(1) = 0.

p(1) = k x (1)² – √2 x 1 + 1

0 = k - √2 + 1

0 = -√2 + 1 + k

k = √2 - 1

(iv) p(x) = kx² – 3x + k, k = ?

Given that x - 1 is a factor of p(x).

So, x - 1 = 0

x = 1

And p(1) = 0.

p(1) = k x (1)² - 3 + k

0 = k - 3 + k

0 = -3 + 2k

2k = 3

k = 3/2