Question

Hi Friends !!
if alpha and beta are the zeroes of the polynomial p(x)=x²-5x+6,find the value of alpha⁴beta²+alpha²beta⁴​ .

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Answer 1

Given:-

→ Polynomial : P(x) = x² - 5x + 6

To find:-

• α⁴β² + α²β⁴

Solution:-

We will find zeros of given polynomial by factorization method:-

→ x² - 5x + 6 = 0

→ x² - 3x - 2x + 6 = 0

→ x(x - 3) - 2(x - 3) = 0

→ (x - 3)(x - 2) = 0

→ x = 3 & x = 2

Therefore,

Zeros of polynomial are 3 & 2 .

Hence,

• α = 3
• β = 2

Now, α⁴β² + α²β⁴

→ (3)⁴ × (2)² + (3)² × (2)⁴

→ (81 × 4) + (9 × 16)

→ 324 + 144

→ 468

Therefore,

Required value = 468 .

Answer 2

Aɴꜱᴡᴇʀ

$\huge\sf{468}$

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Gɪᴠᴇɴ

$\sf{p(x) = {x}^{2} - 5x + 6}$

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ᴛᴏ ꜰɪɴᴅ

${ \alpha }^{4} { \beta }^{2} + { \alpha }^{2} { \beta }^{4}$

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Sᴛᴇᴘꜱ

$\sf{}p(x) = {x}^{2} - 5x + 6 \\ \sf{} = {x }^{2} - 2x - 3x + 6 \\ \sf{}x(x - 2) - 3(x - 2) \\ \sf{}(x - 3)(x - 2) \\ \sf{}x = 3 \: and \: x = 2$

So the zeros of the polynomial are 3 and 2

$\sf{}so \: \alpha = 3 \\ \sf{} \beta = 2 \\ \sf{}so \: sub \: this \: in \: { \alpha }^{4} { \beta }^{2} + { \alpha }^{2} { \beta }^{4}$

$\sf{}we \: get \: ( {3)}^{4} ( {2)}^{2} + {(3)}^{2} {(2)}^{4} \\ \sf{} = (81 \times 4) +( 9 \times 16) \\ \sf{} = 468$

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