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3 years ago

Q(x)=x^3+ax^2+2b is divided by (x-3) and has a remainder of 4. Find the values of "a" and "b"

Mathematics

1 answer:

Sati [7]3 years ago

4 0

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:

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3 years ago

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Likurg_2 [28]

3x + 1 = y
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To solve this system of equations, we are going to use the substitution method. Substitution the equation where the variable is isolated into the second equation. In this system of equations, y is isolated, so we will replace y in the second equation with 3x + 1.

2x + 3y = 14
2x + 3(3x + 1) = 14
2x + 9x + 3 = 14
We will add the like terms and subtract 3 from both sides of the equation.
11x + 3 = 14
11x = 11
x = 1
In this system of equations, x is equal to 1. Now we will go back and solve for y, plugging in 1 for x.
3(1) + 1 = y
2(1) + 3y = 14

3 + 1 = y
2 + 3y = 14

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The solution to this system of equations is (1, 4).

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3 years ago

Q(x)=x^3+ax^2+2b is divided by (x-3) and has a remainder of 4. Find the values of "a" and "b"​

Q(x)=x^3+ax^2+2b is divided by (x-3) and has a remainder of 4. Find the values of "a" and "b"