Question

verify that 2,1 and 1 are the zeroes of the cubic polynomial p(x)=x³-4x²+5x-2 and then verify between the zeroes and the coefficients of the polynomial​

Answer 1

Solution :

p(x) = x³ - 4x² + 5x - 2

$\boxed{\begin{array}{cccc}\sf At\:x = 2&\sf At\: x = 1 &\sf At\:x = 1\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf p(2) = (2)^3 - 4(2)^2 + 5(2) - 2&\sf p(1)=(1)^3 - 4(1)^2 + 5(1) - 2 &\sf p(1) = (1)^3 - 4(1)^2 + 5(1) - 2 \\\\\sf 8 - 4(4) + 10 - 2 &\sf 1 - 4 + 5 - 2&\sf 1+ 4 + 5 - 2 \\\\\sf 18 - 18&\sf 5 - 5&\sf 5 - 5 \\\\\sf = 0 &\sf = 0 &\sf = 0\end{array}}$

Hence, 2,1,1 is the zero of p(x).

Verifying the relationship between zeros and coefficient.

$\boxed{\begin{minipage}{6.5 cm}\underline{\text{For a cubic polynomial :}}\\ \\ \sf p(x) = ax^3 + bx^2 + cx + d\\ \\ \sf Where \:zeroes\: \alpha \:\beta and \gamma \\ \\ \sf \alpha + \beta + \gamma = \dfrac{-b}{a} \\ \\ \sf \alpha\beta+ \beta \gamma+ \gamma\alpha=\dfrac{c}{a} \\ \\\sf sf \alpha \beta \gamma = \dfarc{-d}{a}\end{minipage}}$

For p(x) = x³ - 4x² + 5x - 2

a = 1, b = -4, c = 5, d = -2

and zeroes are,

α = 2, β = 1, γ = 1

So,

1) α + β + γ

➟2 + 1 + 1

➟4

➟ - ( - 4)/1

➟ - b/a

2) αβ + βγ + γα

➟(2)(1) + (1)(1) + (1)(2)

➟5

➟5/1

➟c/a

3) αβγ

➟(2)(1)(1)

➟2

➟- ( - 2)/1

➟-d/a

Hence, Verified.

Answer 2

Answer:

Answer:-

• Given:-

Zeroes of a cubic polynomial are = 2, 1, 1

Cubic polynomial = p(x) = x³ - 4x² + 5x - 2

• Solution:-

i] Comparing the given polynomial with ax³ + bx² + cx + d ,

we get,

a = 1

b = -4

c = 5

d = -2

Further,

Given that,

2 , 1 ,1 are zeroes of the given polynomial

HENCE

p(2) = 1 × (2)³ + (-4) (2)² + 5 (2) + (-2)

→ 8 + (-4 × 4) + 10 - 2

→ 8 - 16 + 10 - 2

→ 18 - 18

→ 0

Now,

p(1) = (1)³ + (-4) (1)² + 5(1) + (-2)

→ 1 + (-4 × 1) + 5 - 2

→ 1 - 4 + 5 - 2

→ 6 - 6

→ 0

Therefore, [2 , 1 ,1 ] are the zeroes of the given polynomial.

ii] So,

We take $\bf{\alpha = 2 , \beta = 1 , \gamma = 1}$

Now,

$\alpha + \beta + \gamma$ = 2 + 1 + 1

→ 5 = $\bf{\dfrac{-b}{a}}$

$\alpha \beta + \beta \gamma + \gamma \alpha$ = 2 × 1 + 1 × 1 + 1 × 2

→ 2 + 1 + 2

→ 5 = $\bf{\dfrac{c}{a}}$

$\alpha\beta\gamma$ = 2 × 1 × 1

→ 2 = $\bf{\dfrac{-d}{a}}$

HENCE VERIFIED