Question

-3,0,2 are the zeroes of the polynomial p (x)=x cube +(a-1)a square +bx+c then find a and c

Answer 1

Answer:

Step-by-step explanation:

p(x)=x^3+(a-1)x^2+bx+c

p(-3)=(-3)^3+(a-1)(-3)^2+b(-3)+c=0

-27+9(a-1)-3b+c=0

-27+9a-9-3b+c=0

9a-3b+0-18=0

9a-3b+18=0----(1)

p(0)=(0)^3+(a-1)0+b(0)+c=0

c=0

p(2)=(2^3)+(a-1)(2^2)+2b+c=0

8+4(a-1)+2b=0

8+4a-4+2b=0

4a+2b+4=0--(2)

solve (1) and (2)

multiply (1) with 2 and (2) with 3

18a-6b+36=0

12a+6b+12=0

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30a=--48

a=-16/10=-1.6

substitute in (2)

4*-1.6+2b+4=0

-6.4+4+2b=0

-2.4=-2b

b=1.2

Answer 2

Answer:

given solution comparison ax^+bx^+cx+c

1. sum of roots alpha + beta + Gamma is = -b/ a
2. -3+0+2=-(a-1)/1
3. -1=-a-1
4. -2=-a
5. a=2