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​

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Answer 1

Aɴꜱᴡᴇʀ

$\huge\sf{\fbox{\fbox{468}}}$

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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:

★$\alpha \:and \:\beta$are the zeroes if the polynomial.

i.e. p(x) =$x^{2}-5x+6$

TO FIND:

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

CONCEPT USED:

★We can find it's zeroes by factorisation using middle term splitting.

★We can also find it's zeroes using standard formula

i. e. $\dfrac{-b±\sqrt{b^{2}-4ac}}{2a}$

of the equation in $ax^{2}+bx+c$ form

ANSWER:

We would be using middle term splitting,

=>$x^{2}-5x+6=0$

=>$x^{2}-3x-2x+6=0$

=>$x(x-3)-2(x-3)=0$

=>$(x-2)(x-3)=0$

Hence x = 3,2

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Now let,

$\alpha= 2\:\:and\:\:\beta=3$

So,

$\alpha^{2}=2^{2}=4$;

$\alpha^{4}=2^{4}=16$

&

$\beta^{2}=3^{2}=9$;

$\beta^{4}=3^{4}=81$

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Therefore,

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

=$16×9+4×81$

=$144+324$

=$468$

Hence $\alpha^{4}\beta^{2}+\alpha^{2}\beta^{4}=468$

$\huge\orange{\boxed{\alpha^{4}\beta^{2}+\alpha^{2}\beta^{4}=468}}$