Question

If the zeros of the polynomial/x) =x*-3x-6x +8 are of the form a-b, a,a + b, find all the zeros.​

Answer 1

✤ᴄᴏʀʀᴇᴄᴛ Qᴜᴇꜱᴛɪᴏɴ :-

If the zeros of the polynomial p(x) =x³-3x²-6x +8 are of the form a-b, a,a + b, find all the zeros.

✤ɢɪᴠᴇɴ:-

p(x) =x³-3x²-6x +8

Let α = a - b

β = a

γ = a + b

✤ꜱᴏʟᴜᴛɪᴏɴ :-

p(x) =x³-3x²-6x +8

$\bf \alpha + \beta + \gamma = \frac{ - (cofficient \: of \: {x}^{2} )}{coefficient \: of \: {x}^{3} }$

$\bf \to a - b + a + a + b = \frac{ - ( - 3)}{1}$

$\bf \to a + a + a = 3$

$\bf \to 3a = 3$

$\bf \to a = \frac{3}{3}$

$\bf \to a = 1$

$\bf \alpha \times \beta \times \gamma = \frac{ - (constant \: term)}{coefficient \: of \: {x}^{3} }$

$\bf \to (a - b) \times a \times (a + b) = \frac{ - 8}{1}$

$\bf \to (a - b) \times (a + b) \times a= - 8$

$\bf \to ( {a}^{2} - {b}^{2} ) \times a = - 8$

$\bf \to {a}^{3} - {b}^{2} a = - 8$

$\bf putting \: a = 1$

$\bf \to {1}^{3} - {b}^{2} \times 1 = - 8$

$\bf \to 1 - {b}^{2} = - 8$

$\bf \to { - b}^{2} = - 8 - 1$

$\bf \to { - b}^{2} = - 9$

$\bf \to {b}^{2} = 9$

$\bf \to b = \sqrt{9}$

$\bf \to b = 3$

Therefore,

a = 1 and b = 3

• α = a - b = 1 - 3 = -2
• β = a = 1
• γ = a + b = 1 + 3 = 4