Question

find the zeros of cubic polynomial x cube minus 5 x square - 2 X + 24 when it is given that the product zeros is 12​

Answer 1

Correct Question:

Find the zeros of the polynomial f(x)=x³- 5x²-2x+ 24 if it is Given that product of its two zeros is 12.

AnswEr:

• x³ - 5x² - 2x + 24

Product of its zeroes is 12.

Let us Consider that the Zeroes of the polynomial be $\sf\:\alpha,\beta\; and \; \gamma$

$\dag\:\:\bold{\underline{\sf{\red{According\:to\:Question\;Now,}}}}$

$\longrightarrow\sf\:\alpha + \beta = 12$ ----Eq(1)

We know that,

$\longrightarrow\sf\: \alpha\:\beta\:\gamma = \dfrac{-Costant \: Term}{Coefficient\: of \: x^2}$

$\longrightarrow\sf\: -24$ ------Eq(2)

And,

$\longrightarrow\sf\:\alpha \:+\:\beta\:+\:\gamma = \dfrac{- Coefficient\: of \:x^2}{Coefficient\: of\: x^3}$

$\longrightarrow\sf\: -\dfrac{(-5)}{1}$

$\longrightarrow\sf\: 5$ --------Eq(3)

$\rule{120}2$

Now, Substituting the Value of $\sf\:\alpha\:\beta$ in Equation (2).

$\longrightarrow\sf\:\alpha \:\beta\:\gamma = -24$

$\longrightarrow\sf\:12 \gamma =- 24$

$\longrightarrow\sf\: \gamma =\dfrac{-24}{\:\:12}$

$\longrightarrow\large{\underline{\boxed{\sf{\red{\gamma\:=\: -2}}}}}$

Now, Substituting the value of $\sf\gamma$ in Equation (3)

$\longrightarrow\sf\:\alpha \:+\:\beta \:+\:\gamma = 5$

$\longrightarrow\sf\:\alpha + \beta + (-2) = 5$

$\longrightarrow\sf\:\alpha + \beta = 5 + 2$

$\longrightarrow\sf\:\alpha + \beta = 7$ -------Eq(4)

$\dag\:\:\small\bold{\underline{\sf{\red{Squaring\: both \:sides}}}}$

$\longrightarrow\sf\:(\alpha + \beta)^2 \:=\: 7^2$

Using Identity

$\sf\:(\alpha + \beta) = (\alpha \:-\beta)^2 + 4 (\alpha\:\beta)$

So,

$\longrightarrow\sf\: (\alpha\:-\:beta )^2 + 4 \times \: 12 = 49$

$\longrightarrow\sf\:(\alpha \:-\:\beta)^2 + 48 = 49$

$\longrightarrow\sf\:(\alpha \:-\:\beta)^2 = 49 - 48$

$\longrightarrow\sf\:(\alpha \:-\:\beta)^2 = 1$

Here, we get $\sf\:\alpha\:-\:\beta = 1$ -------Eq(5)

Adding Equations (4) & (5)

$\longrightarrow\sf\:\alpha + \beta = 7$

$\longrightarrow\sf\:\alpha - \beta = 1$

$\longrightarrow\sf\:2\alpha = 8$

$\longrightarrow\sf\:\alpha = \dfrac{8}{2}$

$\longrightarrow\large{\underline{\boxed{\sf{\pink{\alpha \:=\: 4}}}}}$

$\rule{120}2$

Now, Substituting the Value of $\sf\alpha$ in Equation (5)

$\longrightarrow\sf\: \alpha - \beta = 1$

$\longrightarrow\sf\:4 - \beta = 1$

$\longrightarrow\sf\: - \beta = 4 - 1$

$\longrightarrow\sf\: -\beta = - 3$

$\longrightarrow\large{\underline{\boxed{\sf{\pink{\beta \:=\: 3}}}}}$

$\bold{\underline{\sf{\blue{Hence,\: The\:Zeroes\:are\: 4,\:3,\:-2.}}}}$

Answer 2

$\mathfrak{\huge{\pink{ANSWER}}}$