Question

show that the sum of the zeroes, product
of the zeroes and sum of product of
zeroes taken two at a time of the
to 2x²–5x²-14x + 8 ore 5/2, -4,-7
find zeroes of the polynomial​

Answer 1

Step-by-step explanation:

from the polynomial,

a=2,b=-5,c=-14,d=8

sum of zeroes=-b/a=5/2

product of zeroes=-d/a=-8/2=-4

sum of product of zeroes two taken at a time=c/a=-14/2=-7

the zeroes of the polynomial can be found by synthetic division method

Answer 2

Answer:

Sum of the zeroes ⇒ 5/2

Product of the zeroes ⇒  -4

Product of zeroes taken two each ⇒  -7

Zeroes ⇒ -2, 1/2, 4.

Step-by-step explanation:

We know that,

Sum of zeroes = α + β + γ = -b/a

Product of zeroes = αβγ = -d/a

Product of zeroes taken two at a time = αβ + βγ + γα = c/a

Where,

'a' ⇒ Co-efficient of x³

'b' ⇒ Co-efficient of x²

'c' ⇒ Co-efficient of x

'd' ⇒ Constant term.

Now, We'll substitute and solve.

p(x) = 2x³ - 5x² - 14x + 8

Sum of Zeroes.

$\Longrightarrow \ \mathsf {Sum \ of \ Zeroes = \dfrac{-b}{a}}$

$\Longrightarrow \ \mathsf {Sum \ of \ Zeroes = \dfrac{-(-5)}{2}}$

$\Longrightarrow \ \mathsf {Sum \ of \ Zeroes = \dfrac{5}{2}}$

Product of Zeroes.

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes = \dfrac{-d}{a}}$

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes = \dfrac{-8}{2}}$

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes = -4}$

Product of zeroes taken two at a time.

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes \ taken \ two \ at \ a \ time = \dfrac{c}{a}}$

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes \ taken \ two \ at \ a \ time = \dfrac{-14}{2}}$

$\Longrightarrow \ \mathsf {Product \ of \ Zeroes \ taken \ two \ at \ a \ time = -7}$

To find the zeroes, we can assume x to be a number, and substitute and check if we get 0. If we do, we can assume it to be one of the zeroes, and find the others.

p(x) = 2x³ - 5x² - 14x + 8

Let us assume x to be 2.

p(2) = 2(2)³ - 5(2)² - 14(2) + 8

p(2) = 2(8) - 5(4) - 28 + 8

p(2) = 16 - 20 - 20

p(2) = 16 - 40

p(2) = -24

p(2) ≠ 0

Hence +2 is not a zero of the given polynomial.

Now, Let us assume x to be -2.

p(-2) = 2(-2)³ - 5(-2)² - 14(-2) + 8

p(-2) = 2(-8) - 5(4) + 28 + 8

p(-2) = -16 - 20 + 36

p(-2) = -36 + 36

p(-2) = 0

Hence, -2 is a zero of the given polynomial.

As α, β and γ are the zeroes, we can take α to be -2.

α = -2

With the value of α, we can find β and γ.

Sum of Zeroes = α + β + γ = 5/2

⇒ α + β + γ = 5/2

⇒ -2 + β + γ = 5/2

⇒ β + γ = 5/2 + 2

⇒ β + γ = {5 + 4}/2

⇒ β + γ = 9/2 ...........Eq(1)

Product of zeroes = αβγ = -4

⇒ αβγ = -4

⇒ -2βγ = -4

⇒ βγ = -4/-2

⇒ βγ = 2

⇒ γ = 2/β ...........Eq(2)

Substitute Eq(2) in Eq(1)

$\Longrightarrow \ \sf \beta + \gamma = \dfrac{9}{2}$

$\Longrightarrow \ \sf \beta + \dfrac{2}{\beta} = \dfrac{9}{2}$

$\Longrightarrow \ \sf \dfrac{\beta^2 + 2}{\beta} = \dfrac{9}{2}$

⇒ 2β² + 4 = 9β

⇒ 2β² - 9β + 4 = 0

⇒ 2β² - 8β - β + 4 = 0

⇒ 2β(β - 4) -1(β -4)

⇒ (2β - 1)(β - 4)

Hence the other two zeroes are 1/2 and 4.

β = 4, γ = 1/2

Zeroes

α = -2

β = 1/2

γ = 4