Question

if the zeroes of the polynomial p (x) = x^3 - 9x^2 + 23x - 15 are a-d, a, a+d. find the values of a and d. also find all the zeroes of the polynomial p (x).

Answer 1

X3-9x2+23x-15
= sum of zeros = A+B+C (alpha+beta+gamma)
= -b/a = a-d+a+a+d
= 9= 3a
= a= 3
= product of zeros = A×B×C
= -d/a =(a-d) (a+d) a
= 15 = (a2-d2)3
= 15 = (9-d2)3
= 15= 27-3d2
= 3d2= 12
= d2 = 4
= d= +minus2
= then the zeros are 1,5,3.

Answer 2

Answer:

⇒ zeros of the P(x) = x³ - 9x² + 23x - 15 are  1, 2, 5

Step-by-step explanation:

Given Polynomial P(x) = x³ - 9x² + 23x - 15  

a - d, a and a + d are the roots of P(x)  

here we need to find zeros of the P(x)  

⇒ if  P(y) = Ax³ + Bx² + Cx + D and $\alpha$  β, λ are be the roots then  

sum of the roots α + β + λ = $- \frac{B}{A}$  

product of the roots = αβγ =  $\frac{C}{A}$  

⇒ P(x) = x³ - 9x² + 23x - 15 with Ax³ + Bx² + Cx + D

⇒ A = 1, B = - 9, C= 23 and D =  - 15

⇒ sum of the roots a - d + a + a + d = $-\frac{B}{A}$  = -[-9/1] = 9

                                          3a = 9

                                            a = 3

⇒ (a - d) (a) (a + d) = $\frac{C}{A}$ = -15/1

⇒ ( a² - d²) (a) = - 15

⇒ ( 9 - d²) (3)  = - 15

⇒ 9 - d² = - 5  

⇒ - d² = -4

⇒   d = 2  

a - d = 3 - 2 = 1    and  a+d = 3 + 2 = 5

⇒ zeros of the P(x) = x³ - 9x² + 23x - 15 are  1, 2, 5