Question

Verify that 3 , -1 , -1 /3 are the zeroes of the cubic polynomial

Answer 1

Given:-

•α(Alpha)=3

•β(Beta)=-1

•γ(Gamma)=-⅓

$\rule{200}{1}$

To Find:-

✪The Cubic Polynomial and Verify its Zeros?

$\rule{200}{1}$

AnsWer:-

▼Using Cubic Formula▼

↝k[x³-(α+β+γ)x²+(αβ+βγ+γα)x-(αβγ)]

•α+β+γ=3+(-1)+(-⅓)

•α+β+γ=3-1-⅓

•α+β+γ=2-⅓→[Take LCM]

•α+β+γ=$\frac{6-1}{3}$

•α+β+γ=$\frac{5}{3}$–(1)

•αβ+βγ+γα=3×(-1)+(-1)×(-⅓)+(-⅓)×3

•αβ+βγ+γα=-3+1+(-1)

•αβ+βγ+γα=-2-1

•αβ+βγ+γα=-3–(2)

•αβγ=3×-1×-⅓

•αβγ=1–(3)

✪Using (1),(2)&(3) in the Formula✪

↝k[x³-($\frac{5}{3}$x²+3x-1]

★Let k=3★

↝3[x³-($\frac{5}{3}$x²+3x-1]

↝3x³-5x²+9x-3

☞3x³-5x²+9x-3 is the Cubic Polynomial.

♦As The Zeros Form the Cubic Polynomial,It is verified they are the zeros of Cubic Polynomial♦

$\rule{200}{2}$

Answer 2

Answer:

Given:-

•α(Alpha)=3

•β(Beta)=-1

•γ(Gamma)=-⅓

$$\rule{200}{1}$$

To Find:-

✪The Cubic Polynomial and Verify its Zeros?

$$\rule{200}{1}$$

AnsWer:-

▼Using Cubic Formula▼

↝k[x³-(α+β+γ)x²+(αβ+βγ+γα)x-(αβγ)]

•α+β+γ=3+(-1)+(-⅓)

•α+β+γ=3-1-⅓

•α+β+γ=2-⅓→[Take LCM]

•α+β+γ=$$\frac{6-1}{3}$$

•α+β+γ=$$\frac{5}{3}$$ –(1)

•αβ+βγ+γα=3×(-1)+(-1)×(-⅓)+(-⅓)×3

•αβ+βγ+γα=-3+1+(-1)

•αβ+βγ+γα=-2-1

•αβ+βγ+γα=-3–(2)

•αβγ=3×-1×-⅓

•αβγ=1–(3)

✪Using (1),(2)&(3) in the Formula✪

↝k[x³-($$\frac{5}{3}$$ x²+3x-1]

★Let k=3★

↝3[x³-($$\frac{5}{3}$$ x²+3x-1]

↝3x³-5x²+9x-3

☞3x³-5x²+9x-3 is the Cubic Polynomial.

♦As The Zeros Form the Cubic Polynomial,It is verified they are the zeros of Cubic Polynomial♦

$$\rule{200}{2}$$