Question

2.a) Verify that - 2, 1 and 3 are the zeroes of the cubic polynomial
X^3 - 2x² - 5x + 6 and check the relationship between zeroes and the
co-efficients.​

Answer 1

Hey mate, Here's the answer. . .

Given

${x}^{3} - 2 {x}^{2} - 5x + 6$

substituting -2

$= { - 2}^{3} - 2 ({ - 2})^{2} - 5( - 2) + 6 \\ = - 8 - 8 + 10 + 6 \\ = 16 - 16 = 0$

similarly we can prove for the other numbers.

Now

$sum \: of \: roots = \frac{ - b}{a} = \frac{ - ( - 2)}{1} = 2$

$product \: of \: roots = \frac{c}{a} = \frac{ - 5}{1} = - 5$

$sum \: of \: product \: of \: roots \: taken \: two \: at \: a \: time = \frac{ - d}{a} = \frac{ - 6}{1} = - 6$

Hope it helps! ! ! !

Mark my answer as brainliest! ! ! !

Regards,

Abhinay.

Answer 2

Answer:

p(x)=x^3-2x^2-5x+6

p(-2)=-2^3-2(-2)^2-5(-2)+6

=-8-8+10+6

=0

p(1) = (1)^3-2(1)^2 -5(1)+6

= 1-2-5+6

=0

p(3) = (3)^3-2(3)^2-5(3)+6

=27-18-15+6

=33-33 =0

sum of zeroes =-2+1+3=2

sum of zeroes =-b/a=-(-2)/1=2

product of zeroes =-2*1*3= -6

product of zeroes = -c/a =-6

sum of products of zeroes = (-2*1)+(1*3)+(3*-2)

-2+3-6=-5

sum of products of zeroes=d/a =-5