Consider the cubic function f(x) = x^3 + ax^2 + bx + 4 where (a) and (b) are constants. when f(x) is divided by x-3, the remainder is 10. when f (x) is divided by x+1, the remainder is 6.
show that a = -1 and b = -4

see explanation

Step-by-step explanation:

Using the Remainder theorem

If f(x) is divided by (x - h) then f(h) = remainder, thus

f(3) = 3³ + a(3)² + b(3) + 4 = 10, that is

27 + 9a + 3b + 4 = 10

9a + 3b + 31 = 10 ( subtract 31 from both sides )

9a + 3b = - 21 → (1)

f(- 1) = (- 1)³ + a(- 1)² + b(- 1) + 4 = 6, that is

- 1 + a - b + 4 = 6

a - b + 3 = 6 ( subtract 3 from both sides )

a - b = 3 → (2)

Rearrange (2) expressing a in terms of b

a = 3 + b → (3)

Substitute a = 3 + b into (1)

9(3 + b) + 3b = - 21 ← distribute and simplify left side

27 + 9b + 3b = - 21

12b + 27 = - 21 ( subtract 27 from both sides )

12b = - 48 ( divide both sides by 12 )

b = - 4

Substitute b = - 4 into (3)

a = 3 - 4 = - 1

Hence a = - 1 and b = - 4

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.