Dear all,

Can anyone show me the solution?

Thanks

## 9768 #65

### Re: 9768 #65

For 65 you have:

p(-3): 27-9a+3b=c

p(2): -8-4a-2b=c

Equating the two, it simplifies down to 7+b=a.

We also have that p'(-3)<0; so 3x^2+2ax+b<0 ==> 27-6a=b<0 ==> 27+b<6a. Substituting in for a, 27+b<6(7+b)==> -15<5b ==> -3<b.

Now lets look at -8-4a-2b=c. We have that b>-3 ==> -2b <-6 ==> -2b <-14. Similarly, a>2 ==> -4a <-8. Adding the two we have -2b-8-4a < -22 ==> c <-22. Thus -27 is the only number that satisfies.

Hopefully this helps.

p(-3): 27-9a+3b=c

p(2): -8-4a-2b=c

Equating the two, it simplifies down to 7+b=a.

We also have that p'(-3)<0; so 3x^2+2ax+b<0 ==> 27-6a=b<0 ==> 27+b<6a. Substituting in for a, 27+b<6(7+b)==> -15<5b ==> -3<b.

Now lets look at -8-4a-2b=c. We have that b>-3 ==> -2b <-6 ==> -2b <-14. Similarly, a>2 ==> -4a <-8. Adding the two we have -2b-8-4a < -22 ==> c <-22. Thus -27 is the only number that satisfies.

Hopefully this helps.

### Re: 9768 #65

3rd degree polynomials with first coefficient 1, and 2 known roots

c is the y i tercept, or p(0)

Going to the factorial decomposition of p

p = (x+3)(x-2)(x-k)

Plugging 0

c = 6k

Checking that p is decreasing in -3, coming from positive values, and unbounded in -oo, we observe that the last root k is less than -3

Hence, c <-18

c is the y i tercept, or p(0)

Going to the factorial decomposition of p

p = (x+3)(x-2)(x-k)

Plugging 0

c = 6k

Checking that p is decreasing in -3, coming from positive values, and unbounded in -oo, we observe that the last root k is less than -3

Hence, c <-18