Question

Verify that 3, -2,1 are the zeros of the cubic polynomial
p(x) = x^3 - 2x^2 - 5x +6 and verify the relation between its zeros and
coefficients.​

Answer 1

Answer:

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

x=3,-2,1

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

=27-2(9)-15+6

=27-18-15+6

=33-33

=0(It is a factor)

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

=-8-2(4)-(-10)+6

=-8-8+10+6

=16-16

=0(It is a factor)

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

=1-2-5+6

=7-7

=0(It is a factor)

RELATION BETWEEN ITS ZEROES AND COEFFICIENT:

$\alpha = 3. \beta = - 2. \gamma = 1$

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \alpha \beta \gamma = \frac{ - d}{a}$

ax^3+bx^2+cx+d=x^3-2x^2-5x+6

a=1;b=-2;c=-5;d=6

(3)+(-2)+(1)=-(-2)/1

4-2=2/1

2=2

$\alpha + \beta + \gamma = \frac{ - b}{a}$

(3)(-2)+(-2)(1)+(1)(3)=-5/1

-6-2+3=-5

-8+3=-5

-5=-5

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

(3)(-2)(1)=-6/1

-6=-6

$\alpha \beta \gamma = \frac{ - d}{a}$

HOPE IT HELPS YOU....