Can someone pls help me with this. The polynomial f(x) 3x^3+ax^2+bx+16 has a double root at X=2.?

Question

find the values of a and b.thank you

Answer 1

I think John T got the right answer because the degree is one more so you can pretty much guess the missing factor and distibute. But if that wasn't possible the method is to equate f(2)=0 and f'(2)=0 where f'(x)=derivative of f w/r/to x=9x^2+2ax+b

So you end up with

0=f(2)= 3*2^3+a*2^2+ b*2+16
and
0=f'(2)= 9*2^2+2*a*2+b

So you then have 2 equations in two unknows and you use the elimination process to solve for a and b.

Answer 2

So if 3x^3+ax^2+bx+16 has a double root at x=2, where the expression evaluates to zero

So, 24 + 4a+2b +16 = 0

4a+2b = -40

2a+b = -20

Let's look at the numbers. The last component has to be (3x +4)
since that is the only thing that will have the ending numbers (3x^3) and 16.

multiply (x-2)^2 and (3x+ 4)

I get -8 for a and -4 for b, which matches my equation.

Answer 3

6.0-