Answer:
Let p(x) = x3 + ax2 + bx +6
(x-2) is a factor of the polynomial x3 + ax2 + b x +6
p(2) = 0
p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0
7 +2 a +b = 0
b = - 7 -2a -(i)
x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3
11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)
Equating the value of b from (ii) and (i) , we have
(- 7 -2a) = (-10 - 3a)
a = -3
Substituting a = -3 in (i), we get
b = - 7 -2(-3) = -7 + 6 = -1
Thus the values of a and b are -3 and -1 respectively.
Step-by-step explanation:
Solution:
1) Simplify \frac{1}{6}x to \frac{x}{6}
y=\frac{x}{6}-2
2) Add 2 to both sides
y+2=\frac{x}{6}
3) Multiply both sides by 6
(y+2)\times 6=x
4) Regroup terms
6(y+2)=x
5) Switch sides
x=6(y+2)
Done!
8333/5000 have a brainly day :)
Answer:what is this
Step-by-step explanation: the answers are too small
