Question

if alpha and beta are zeros of p(x)
$x {}^{2} - 2x - 8$
Find
$\alpha {}^{3} - \beta {}^{3}$
​

Answer 1

Answer:

Given equation :- x² + 4x + 3

Factorising it by middle term splitting :-

x² + 4x + 3

x² + 3x + x + 3

x ( x + 3 ) + 1 ( x + 3 )

( x + 3 ) ( x + 1 )

• ( x + 3 ) = 0

x = ( - 3 )

• ( x + 1 ) = 0

x = ( - 1 )

• The zeros of new equation are :-

So, the Zeros of new equation are 2 and 0

• Sum of the Zeros are :-

0 + 2 = 2

• Product of the Zeros are :-

0 × 2 = 0

♯ To form the quadratic equation we have formula as :-

x² - ( sum of Zeros )x + (product of Zeros)

Putting value in it !!

x² - 2x + 0

So, the required quadratic equation is

x² - 2x .

Answer 2

Aɴꜱᴡᴇʀ

$\large \tt{\red{ { \alpha }^{3} - { \beta }^{3} = 72}}$

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Gɪᴠᴇɴ

$\large \tt{}p(x) = {x}^{2} - 2x - 8\\ \\ \large \tt{} \alpha \: and \: \beta \: are \: the \: zeros \: of \:p ( x)$

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ᴛᴏ ꜰɪɴᴅ

$\large \tt { \alpha }^{3} - { \beta }^{3}$

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Sᴛᴇᴘꜱ

Given the values of p(x) then we have to simply find its zeros

x²-2x-8 can also be written as

$\large \tt \dashrightarrow {} {x}^{2} - 4x + 2x - 8 \\ \\ \large \tt{}taking \:common \: out \\ \\ \large \tt{}x(x - 4) + 2(x - 4) \\ \\ \large \tt{}zeros \: are \: \green{(x - 4)} \orange{(x + 2)} \\ \\ \large \tt{} \alpha = 4 \: \: \: and \: \: \: \beta = - 2$

$\large \sf{ \bf{ { so \: \: \alpha }^{3} - { \beta }^{3} \: \: \: is \: equal \: to }} \\ \\ \large \tt{} \leadsto(4 {)}^{3} - ( - 2 {)}^{3} \\ \\ \large \tt{} \leadsto(64) - ( - 8) \\ \\ \large \tt {} \leadsto64 + 8 \\ \\ \large \tt{} \pink{ \dashrightarrow72}$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$