Question

Verify that 1,4,7 are the zeros of the poynomialx^3-12x^2+39x-28 .also verify the relationship between the zeros and the coefficient

Answer 1

Heya !!!

X = 1

P(X) = X³ - 12X² + 39X - 28

P(1) = (1)³ - 12 × +(1)² + 39 × 1 - 28

=> (1)³ - 12 × (1)² + 39 × 1 - 28

=> 1 - 12 + 39 - 28

=> -40 + 40

=> 0

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X = 4

P(X) = X³ - 12X² + 39X - 28

=> (4)³ - 12 × (4)² + 39 × 4 - 28

=> 64 - 12 × 16 + 156 - 28

=> 64 - 192 + 156 - 28

=> 64 + 156 - 192 - 28

=> 220 - 220

=> 0

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X = 7

P(X) = X³ - 12X² +39X -28

=> (7)³ - 12 × (7)² + 39 × 7 -28

=> 343 - 12 × 49 + 273 - 28

=> 343 - 588 + 273 - 28

=> 343 + 273 - 588 - 28

=> 616 - 616

=> 0

As we can see that in all the three cases remainder is equal to 0. So 1,4 and 7 are the zeroes of the given polynomial X³-12X²+39X-28.

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RELATIONSHIP BETWEEN THE ZEROES AND COEFFICIENT.

Let ,

Alpha = 1 , Beta = 4 and gama = 7

Sum of zeroes = (Alpha + Beta + Gama ) = ( 1+4+7) = 12/1 = -(Coefficient of X²) /( Coefficient of X³).

Product of zeroes = ( Alpha × Beta × Gama) = ( 1 × 4 × 7) = 28/1 = -(Constant term/Coefficient of X².

★ HOPE IT WILL HELP YOU ★

Answer 2

Hey!!

_________________

Let the zeroes be 1, 4, and 7

At x= 1

p (1) = 1^3 -12 × 1^2 + 39 × 1 - 28

= 1 - 12 + 39 -28

= 0
Hence, 1 is the factor of p (x)

At x = 4

p(4) = 4^3 - 12× 4^2 + 39 × 4 - 28

= 64 - 192 + 156 - 28

= 0

Hence, 4 is the factoe of p (x)

At x= 7

p (7) = 7^3 - 12 × 7^2 + 39 × 7 - 28

= 343 - 588 + 273 - 28

= 0

Hence, 7 is the factor of p (x)

p (x) = x^3 - 12x^2 + 39x - 28

It is in the form p (x) = ax^3 - bx^2 + cx - d

a = x , b= -12 , c = 39 , d = -28

alpha = 1 , beta = 4 , gemma = 7

alpha + beta + gemma = -b/a

=> 1 + 4 +7 = -(-12)/1

=> 12 = 12.

alpha×beta + beta×gemma + gemma×alpha = c/a

=> 1×4 + 4×7 + 7×1 = 39/1

=> 4 + 28 + 7 = 39

=> 39 = 39.

alpha×beta×gemma = d/a

=> 1 × 4 × 7 = -28/1

=> 28 = 28.

Hence the relationship between the zeroes and the coefficient is verified.

Hope it will helps you:-)