Question

find the cubic polynomial, whose zeros are 3, 1/2 and -1​

Answer 1

Given:

Three zeroes which are 3, 1/2 and -1​

To find:

The cubic polynomial whose zeroes are 3, 1/2, and -1​.

So,

Let

$\alpha=3$

$\beta= \frac{1}{2}$

$\gamma = (-1)$

So,

We know that,

If α, β, γ are the zeroes of the cubic polynomial, then the polynomial will be

$k(\boxed{\red{\sf x^{3}-(sum\ of\ zeroes)x^{2}+(product\ of\ zeroes\ two\ taken\ at\ time)x - (product\ of\ zeroes)}})$

Which is

$k(\boxed{\red{\sf x^{3}-(\alpha+\beta+\gamma)x^{2}+(\alpha \beta + \beta \gamma + \gamma \alpha)x - (\alpha \beta \gamma )}})$

Where 'k' is a constant.

So,

Sum of zeroes

= α + β + γ

= $3+\frac{1}{2} + (-1)$

= $\frac{6+1-2}{2}$

= $\frac{7-2}{2}$

= $\frac{5}{2}$

Now,

αβ  +βγ + γα

= $[3*\frac{1}{2}]+[\frac{1}{2}*(-1)] + [(-1)*3]$

= $[\frac{3}{2}]+[\frac{-1}{2}] + [-3]$

= $\frac{3+(-1)-6}{2}$

= $\frac{3-7}{2}$

= $\frac{-4}{2}$

= -2

Also,

αβγ

= $3 * \frac{1}{2}* (-1)$

= $\frac{-3}{2}$

Now,

The cubic polynomial is

$\boxed{\red{\sf x^{3}-(\frac{5}{2})x^{2}+(-2)x - (\frac{-3}{2} )}}$

= $\red{\sf x^{3}-\frac{5}{2}x^{2}-2x +\frac{3}{2} }$

= $\red{\sf \frac{2x^{3}-5x^{2}-4x +3}{2} }$

= 2x³ - 5x² - 4x + 3

It is the required cubic polynomial.

Answer 2

Zeroes of the cubic polynomial-

$\star\sf\ \ \alpha=3\ \ ; \ \star\ \beta=\ ^1\!/_2\ \ ;\ \ \star\ \gamma= (-1)$

Now we know that -

$\scriptsize{\boxed{\sf\ x^3-(\alpha+\beta+ \gamma )x^2 +( \alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)}}$

Now the value of

$\star\sf\ \ \ \ (\alpha+\beta+\gamma)\\ \\ \\\implies\sf \Big[ 3+ \dfrac{1}{2}+(-1)\Big]\\ \\ \\\implies\sf \Big[ \dfrac{6+1-2}{2}\Big]\\ \\ \\\implies{\boxed{\sf \ ^5\!/_2}}$

$\star\sf \ \ \ \ (\alpha\beta+\beta\gamma+\gamma\alpha)\\ \\ \\\implies\sf \bigg[ \Big(3\times \dfrac{1}{2}\Big)+\Big(\dfrac{1}{2}\times (-1)\Big)+\Big(-1\times 3\Big)\bigg]\\ \\ \\\implies\sf \Big[ \dfrac{3}{2}+\dfrac{(-1)}{2}+(-3)\Big]\\ \\ \\\implies\sf \Big[\dfrac{3-1-6}{2}\Big]\\ \\ \\\implies{\boxed{\sf \ ^{-4}\!/_2}}$

Now

$\star\sf\ \ \ \ (\alpha\beta\gamma)\\ \\ \\\implies\sf \Big[3\times( ^1\!/_2)\times (-1)\Big]\\ \\ \\ \implies{\boxed{\sf\ ^{-3}\!/_2}}$

Now the polynomial-

By substituting the values -

$\scriptsize{\sf\ x^3-(\alpha+\beta+ \gamma )x^2 +( \alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)}\\ \\ \\ \implies\sf \Big[x^3-(^5\!/_2)x^2+(^{-4}\!/_2)x-(^{-3}\!/_{2})\Big]\\ \\ \\ \implies\sf \dfrac{(2x^3-5x^2-4x+3}{2}=0\\ \\ \\ \dashrightarrow{\boxed{\purple{\sf2x^3-5x^2-4x+3}}}$