Question

. Verify that 2, -3, and 4 are zeroes of cubic polynomial f(x) = x3 – 3x2 – 10x + 24. Also verify the relation between the zeroes and the coefficients.

Answer 1

Answer:

Hello mate..

$f(x) = {x}^{3?} - 3 {x}^{2} - 10x + 24 \\ then \: verify \:$

2,-3,4

$f(2) = {2}^{3?} - 3 {2}^{2} - 10 \times 2 + 24 \\ = 8 - 12 - 20 + 24 \\ = 0$

So 2 is the zero of the equation..

$f( - 3) = { - 3}^{?3} - 3. { - 3}^{2} - 30 + 24 \\ = - 27 - 27 - 30 + 24 \\ = 0$

So - 3 is also zero of this equation..

$f(4) = {4}^{?3} - {4}^{2} .3 - 10 \times 4 + 24 \\ = 64 - 48 - 40 + 24 = 0$

So 2,-3,4 are the zeroes of equation..

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

=2-3+4=-(-3)/1

=3

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a?}$

2×-3+-3×4+4×2=-10/1

$\alpha \beta \gamma = \frac{ - d}{a?}$

2×-3×4=24/1

So verified that 2,-3,4 are zeroes of the equation..

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