# Show that f(x) = 0 has only one root

1. Jul 6, 2016

### Clever Penguin

1. The problem statement, all variables and given/known data

f(x) = 2x3+ax2+bx+10
When f(x)/(2x-1) the remainder is 12
When f(x)/(x+1) there is no remainder
a) Find the value of a and b
b) Show that f(x) = 0 has only one root

2. Relevant equations

None

3. The attempt at a solution

a) (2x-1)=0
x=1/2

f(1/2) = 12 = 2(1/2)3+b(1/2)2+a(1/2)+10
1/4+a/4+b/2+10=12
1+a+2b+40=48
a+2b=7

(x+1)=0
x=-1

f(-1)= 0 = 2(-1)3+a(-1)2+b(-1)+10
a-b+8=0
b=a+8

b=(7-2b)+8
b=15-2b
3b=15
b=5

a=7-2b
a=7-10
a=-3

a=-3 and b=5

b) 2x3-3x2+5x+10=0
Now I need to factorise this, but I don't know how
I tried using x as a common factor, but its not. 10 does not have a factor of x.

Last edited: Jul 6, 2016
2. Jul 6, 2016

### Delta²

You know x+1 is a factor since the f(-1)=0 or the division by x+1 doesn't leave any reminder.

3. Jul 6, 2016

### Ray Vickson

The formula you $f(x) = 2x^3+ax^2 = bx + 10$ makes no sense. Do you mean $f(x) = 2x^3+ax^2 + bx + 10$ or $f(x) = 2x^3 + ax^2 - bx -10$, or what?

4. Jul 6, 2016

### Clever Penguin

So I use long division?

(2x3-3x2 + 5x + 10)/(x+1)?

I changed it
I meant f(x) = 2x3+ax2 + bx + 10

5. Jul 6, 2016

### Delta²

yes do the polynomial division and you 'll find the quotient polynomial of 2nd degree is such that it doesn't have real roots.

6. Jul 6, 2016

### Clever Penguin

I got (2x3-3x2 + 5x + 10)/(x+1) = 2x2-5x+10

So 2x3-3x2 + 5x + 10 = (x+1)(2x2-5x+10)

7. Jul 6, 2016

### Delta²

Very nice I guess you can see why $2x^2-5x+10$ does not have any real roots... So you original function written as the product of those two polynomials has how many real roots?

8. Jul 6, 2016

One
Thanks