Math, asked by sravin, 7 months ago

if 1 and 2 are two zeroes of the polynomial x^3-6x^2 +ax+b then find the valve of a and b

Answers

Answered by Anonymous
3

Given :-

  • 1 and 2 are two zeroes of polynomial x³-6x²+ax+b.

To find :-

  • Value of a and b.

Solution :-

Since , 1 and 2 are zeroes of the polynomial,

p(1) and p(2) will be 0.

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➾ For p(1) = 0,

 \tt{p(x) =  {x}^{3} - 6 {x}^{2}  + ax + b } \\  \\   \tt{p(1) = {1}^{3}  - 6 {(1)}^{2} + a(1) + b  } \\  \\  \tt{1 - 6 + a + b = 0} \\  \\  \tt{ \red{a + b = 5}} \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: equation \: 1 \\

➾For p(2) = 0

 \tt{p(x) =  {x}^{3}  - 6 {x}^{2}  + ax + b} \\  \\   \tt{p(2) =  {2}^{3}  - 6 {(2)}^{2} + a(2) + b } \\  \\  \tt{8 - 6(4) + 2a + b = 0} \\  \\  \tt{8 - 24 + 2a + b = 0} \\  \\  \tt{ - 16 + 2a + b = 0} \\  \\  \tt{ \red{2a + b = 16} \:  \:  \:  \:  \:  \:  \:  \:  \:  \: equation \:  \: 2}

♣ Substracting equation 1 by equation 2 we get,

➥2a+b-(a+b) = 16-5

➥2a+b-a-b = 11

➥a = 11

➣Putting a = 11 in equation 1 we get,

➥a+b = 5

➥11+b = 5

➥b = -6

∴Value of a is 11 and b is -6.

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