Step 1- Divide the polynomial by x-2 via normal division method. You will get remainder in the form of a and b. ( 2(14+a) + b. Equate it to 2
Step 2- Divide the polynomial by x+2 via normal division method. You will get remainder in the form of a and b. Equate it to -2.
Step 3- Now, you have got 2 equations. Now, since both of these equations are in form of a and b, solve them and you will get the answer for a and b. ( it will be a=-7 and b=-12!)
Hope this helps!
Hello,
Let P(x) = 2x^3+ 3x^2+ ax+ b
Now when the P(x) divides by x-2 then the reminder is b+ 2a+ 28 but according to question reminder should be +2. So the b+ 2a+ 28 = 2
Hence b+ 2a = -26 this is 1 equation
Now when the P(x) divides by x+2 then the reminder is b- 2a- 4 but according to question reminder should be -2 . So the b- 2a- 4= -2
Hence b- 2a = 2 this is 2 equation.
By solving both the equation you may get the values
a = -7
b = -12
Hope this may helps you!
Question : A number x, when divided by 289, leaves 18 as the remainder. The same number when divided by 17 leaves y as a remainder. The value of y is:
Option 1: 5
Option 2: 2
Option 3: 7
Option 4: 1
Question : Let $x$ be the least number which, when divided by 5, 6, 7, and 8, leaves a remainder of 3 in each case, but when divided by 9, leaves no remainder. The sum of the digits of $x$ is:
Option 1: 21
Option 2: 22
Option 3: 18
Option 4: 24
Question : Find the least number which when divided by 12, 18, 24, and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Option 1: 366
Option 2: 364
Option 3: 384
Option 4: 634
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