Question

If the zeroes of polynomial p (x) = x^2 - x - 30 and alpha and beta , find a polynomial with zeroes 2-alpha and 2-beta.​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given: $\alpha$ and $\beta$ are zeros of polynomial

$p(x) = {x}^{2} - x - 30$

To find: Find a polynomial with zeroes $2-\alpha$ and $2-\beta$.

Solution:

Write the relationship of zeros with coefficient of polynomial,if polynomial is $ax^2+bx+c$

$\alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a}$

here for polynomial $x^2-x-30$

$\alpha + \beta = \frac{ - ( - 1)}{1} = 1 ...eq1\\ \\ \alpha \beta = \frac{ - 30}{1} = - 30...eq2$

Let

$\bold{\green{m = 2 - \alpha }} \\ \\\bold{\pink{ n = 2 - \beta }}$

A polynomial having zeros m and n can be expressed as

$\bold{\red{{x}^{2} - (m + n)x + mn}}$

Find m+n:

$m + n = 2 - \alpha + 2 - \beta \\ \\ m + n = 4 - ( \alpha + \beta ) \\ \\ m + n = 4 - 1 \\ \\ m + n = 3...eq3$

Find mn:

$mn = (2 - \alpha )(2 - \beta ) \\ \\ mn = 4 - 2 \beta - 2 \alpha + \alpha \beta \\ \\ mn = 4 - 2( \alpha + \beta ) + \alpha \beta \\ \\ mn = 4 - 2(1) - 30 \\ \\ mn = 4 - 2 - 30 \\ \\ mn = - 28...eq4$

Put the value from eq3 and eq4 in polynomial

${x}^{2} - 3x - 28$

Final answer:

The Polynomial having zeros $2-\alpha$ and $2-\beta$ is

$\bold{\pink{{x}^{2} - 3x - 28}}$

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