Question

Which term of AP: 121 , 117, 113 ........is its first negative term.
. Obtain all zeroes of 3x4 + 6x3 - 2 x2 - 10 x- 5. If two of its zeroes are
and - 5/3
5/3
​

Answer 1

Answer:

(i) A32 (-3)

(ii) Roots of f(x) = √(5/3), -√(5/3), -1, -1

Step-by-step explanation:

In the given A.P,

a = 121

d = 117 - 121 = - 4

an = a + (n-1)d

= 121 - 4(n-1)

We could assume that an = 0, then

121 = 4(n-1) (if an = 0)

121/4 = n-1

30.25 = n - 1 => n = 31.25 ( for An to be zero)

31.25 is approximately 31

So the first negative term is the successor of a31

Therefore, first negative term is (31+ 1) th term

or 32nd term

a32 = 121 - 4(31)

= 121 - 124 = -3

(ii)Given, roots of f(x) = 3x⁴ + 6x³ - 2x²- 10x - 5

are √5/3 & -√5/3

Please note : Roots of this equation are not 5/3 and -5/3, rather, it is √(5/3), and -√(5/3)

Therefore, [x -√(5/3)] & [x + √(5/3)] are factors of f(x)

(x -√5/3) (x + √5/3) = x² - (√5/3)² = x² - 5/3

g(x) = x² - 5/3

f(x) = q(x) • g(x) + r(x)

q(x) = 3x² + 6x + 3 { refer the attached image }

r(x) = 0

q(x) is an quadratic equation. To find it's roots, we use splitting the middle term method.

3x² + 6x + 3

3x² + 3x + 3x + 3.

[ splitting the middle term such that, the product of the digits = 9x² ( 3x² × 3) and sum of the digits = 6x ]

3x(x + 1) + 3(x + 1)

(3x + 3) (x + 1)

3(x + 1)(x + 1)

Therefore, x = -1 ,and x = -1

So, Roots of f(x) = √(5/3), -√(5/3), -1, -1