Question

if a,b,y are the zeros of the polynomial (x cube× 6x square - x + 30) then ( ab + by+ ya) is equal to​

Answer 1

Answer:

αβ + βγ + αγ = (-1)

Step-by-step explanation:

We have,

x³ - 6x² - x + 30 = 0

We are given that,

zeroes of the polynomial are α, β, and γ

We can solve it in two ways

Method 1 (Fast solution)

Now, we know that,

The relation between the zeroes and it's polynomial function is that,

x³ - (α + β + γ)x² + (αβ + βγ + γα)x - (αβγ) = 0

We have the polynomial,

x³ - 6x² - x + 30 = 0

It can also be written as,

x³ - 6x² + (-1)x + 30 = 0

Now, when we compare with the above form we get that,

(αβ + βγ + γα) = (-1)

Method 2 (Long/Trial and error)

We have,

p(x) = x³ - 6x² - x + 30

To get their zeroes, we need p(x) = 0

Let x = 1

p(1) = (1)³ - 6(1)² - 1 + 30

p(1) = 1 - 6 - 1 + 30

p(1) = 24

So,

x ≠ 1

Similarly we can check for x = (-1)

Then, we get,

p(-1) = (-1)³ - 6(-1)² - (-1) + 30

p(-1) = (-1) - 6 + 1 + 30

p(-1) = 24

Then,

x ≠ (-1)

Let's check with x = 2

p(2) = (2)³ - 6(2)² - (2) + 30

p(2) = 8 - 24 - 2 + 30

p(2) = 12

So,

x ≠ 2

Try with x = (-2)

p(-2) = (-2)³ - 6(-2)² - (-2) + 30

p(-2) = (-8) - 24 + 2 + 30

p(-2) = 32 - 32

p(-2) = 0

Hence, we get

x = (-2)

So,

(x + 2) is a factor of p(x)

So,

(x³ - 6x² - x + 30)/(x + 2)

= x² - 8x + 15

So,

x³ - 6x² - x + 30 = (x + 2)(x² - 8x + 15)

Finding zeroes of p(x)

So,

p(x) = (x + 2)(x² - 8x + 15)

Using Splitting the Middle term method,

x² - 8x + 15

ax² + bx + c

Sum = b = (-8)

Product = (a × c) = 15

So, factors are (-3) and (-5)

Thus,

(x + 2)(x² - 8x + 15)

= (x + 2)(x² - 3x - 5x + 15)

= (x + 2)(x(x - 3) - 5(x - 3))

= (x + 2)(x - 3)(x - 5)

Then, we get,

α = (-2)

β = 3

γ = 5

Then,

αβ = (-2) × 3 = (-6)

βγ = 3 × 5 = 15

αγ = (-2) × 5 = (-10)

So,

αβ + βγ + αγ = (-6) + 15 + (-10)

= 15 - 16

= (-1)

Hence,

αβ + βγ + αγ = (-1)

Hope it helped you and believing you understood it...All the best