Question

Verify 1 ,-2and 1/2 are the zeroes of the polynomial 2x^3+x^2-5x+2 and verify the relationship between the zeroes and coefficient?
CLASS 10 POLYNOMIAL​

Answer 1

Step-by-step explanation:

1, -2 and 1/2 will zeroes of polynomial p(x) if p(1)=0, p(-2)=0 and p(1/2)=0

p(x)=2x^3+x^2-5x+2

p(1)=2(1³)+1²-5(1)+2=2+1-5+2

=5-5=0

hence, '1' is a zeroes of p(x)

similarly check p(-2) & p(1/2)

Answer 2

let, f(x) = 2x³ + x² - 5x + 2

1st to verify: \frac{1}{2},1,-2 are zeroes of f(x)

put x=\frac{1}{2}

f(x) = 2(\frac{1}{2})^3+(\frac{1}{2})^2-5\times\frac{1}{2}+2

= 2\times\frac{1}{8}+\frac{1}{4}-\frac{5}{2}+2

= \frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2

= \frac{1+1-10+8}{4}

= 0

Put x = 1

f(x) = 2(1)³ + (1)² - 5(1) + 2

= 2 + 1 - 5 + 2

= 0

put x = -2

f(x) = 2(-2)³ + (-2)² - 5(-2) + 2

= 2 × (-8) + 4 + 10 + 2

= -16 +16

= 0

Therefore, \frac{1}{2},1,-2 are zeroes of given cubic polynomial.

Comparing given cubic equation with standard equation ax³ + bx² + cx + d

we get a = 2 , b = 1 , c = -5 & d = 2

Also take, \alpha=\frac{1}{2}\:,\:\beta=1\:and\:\gamma=-2

Now we verify the relation between zeroes and their coefficient,

\alpha +\beta +\gamma =\frac{1}{2}+1+(-2)=\frac{1+2-4}{2}=\frac{-1}{2}=\frac{-b}{a}

\alpha \,.\,\beta +\beta \,.\,\gamma +\gamma \,.\,\alpha =\frac{1}{2}\times1+1\times(-2)+(-2)\times\frac{1}{2}=\frac{1}{2}-2-1=\frac{1-4-2}{2}=\frac{-5}{2}=\frac{c}{a}

\alpha\,.\,\beta \,.\,\gamma =\frac{1}{2}\times1\times(-2)=-1=\frac{-2}{2}=\frac{-d}{a}

Hence Verified