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Find k so that x² + 2x + k is a factor of 2x⁴ + x³ - 14 x² + 5x + 6. Also find all the zeroes of the two polynomials

Solution:

Given, p(x) = 2x⁴ + x³ - 14 x² + 5x + 6.

g(x) = x² + 2x + k

We have to find the zeros of the polynomial.

The division algorithm states that given any polynomial p(x) and any non-zero

polynomial g(x), there are polynomials q(x) and r(x) such that

p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).

Let r(x) = 0

So, p(x) = g(x) q(x)

By using long division,

Find k so that x² + 2x + k is a factor of 2x⁴ + x³ - 14 x² + 5x + 6. Also find all the zeroes of the two polynomials

So, q(x) = 2x² - 3x - 8 - 2k

r(x) = (21+7k)x + (2k² + 8k + 6)

By comparing the coefficients of (21+7k)x and 2k² + 8k + 6

2k² + 8k + 6 = 0

2(k² + 4k + 3) = 0

k² + 3k + k + 3 = 0

k(k + 3) + (k + 3) = 0

(k + 1)(k + 3) = 0

Now, k + 1 = 0

k = -1

Also, k + 3 = 0

k = -3

So, k = -1, -3.

When k = -1,

21 + 7k = 0

= 21 + 7(-1)

= 21 - 7

= 14

21 + 7k is not equal to zero.

So, k = -1 is neglected.

When k = -3,

21 + 7k = 0

= 21 + 7(-3)

= 21 - 21

= 0

Therefore, the value of k is -3.

Now, g(x) = x² + 2x - 3

x² + 2x - 3 = 0

x² - x + 3x - 3 = 0

x(x - 1) + 3(x - 1) = 0

(x + 3)(x - 1) = 0

Now, x + 3 = 0

x = -3

Also, x - 1 = 0

x = 1

Now, q(x) = 2x² - 3x - 8 - 2k

= 2x² - 3x - 8 - 2(-3)

= 2x² - 3x - 8 + 6

= 2x² - 3x - 2

On factoring,

2x² - 3x - 2 = 0

2x² - 4x + x - 2 = 0

2x(x - 2) + (x - 2) = 0

(2x + 1)(x - 2) = 0

Now, 2x + 1 = 0

2x = -1

x = -1/2

Also, x - 2 = 0

x = 2

We know that g(x) and q(x) are the factors of p(x).

So, the zeros of g(x) and q(x) will be the zeros of p(x).

Therefore, the zeros of p(x) = -3, -1/2, 1 and 2.

✦ Try This: Find the value of k so that x⁴ - 4x³ + 5x² - 2x + k is divisible by x² - 2x + 2

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 2

NCERT Exemplar Class 10 Maths Exercise 2.4 Problem 4

Find k so that x² + 2x + k is a factor of 2x⁴ + x³ - 14 x² + 5x + 6. Also find all the zeroes of the two polynomials

Summary:

The value of k so that x² + 2x + k is a factor of 2x⁴ + x³ - 14 x² + 5x + 6 i s- 3. All the zeroes of the two polynomials are -3, -1/2, 1 and 2

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