Question

Good night friends !! Answer this !!
if alpha and beta are the zeroes of the polynomial p(x)=x²-5x+6,find the value of alpha⁴beta²+alpha²beta⁴​

Answer 1

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

Answer 2

Given:-

Polynomial :-

→ P(x) = x² - 5x + 6

We can find zeros by middle term spitting

→ x² - 5x + 6 = 0

→ x² - 3x - 2x + 6 = 0

→ x(x - 3) - 2(x - 3) = 0

→ (x - 2)(x - 3) = 0

At first,

→ x - 2 = 0

→ x = 2

Again,

→ (x - 3) = 0

→ x = 3

Therefore,

Zeros of polynomial = 2 & 3 .

$\rule{200}{1}$

Here,

• α = 2
• β = 3

Now,we have to find value of :-

• α⁴β² + α²β⁴

Putting all values:-

→ (2)⁴ × (3)² + (2)² × (3)⁴

→ (16 × 9) + (4 × 81)

→ 144 + 324

→ 468

Therefore,

Required value is :— 468 .