# A polynomial function f(x) = ax^3+bx^2-28x+15 is divisible by (x+3) and f(x)+60 is divisible by (x-3). How do you find a, b and solve f(x) = 0 ?

Jun 23, 2015

Derive linear equations for $a$ and $b$ to find $a = 2$, $b = - 5$ and hence:

$f \left(x\right) = \left(2 x - 1\right) \left(x - 5\right) \left(x + 3\right)$

giving roots of $f \left(x\right) = 0$

$x = \frac{1}{2}$, $x = 5$ and $x = - 3$

#### Explanation:

Looking at the coefficients of ${x}^{3}$, ${x}^{2}$ and the constant term,
we find:

[1] $a {x}^{3} + b {x}^{2} - 28 x + 15$

$= \left(a {x}^{2} + \left(b - 3 a\right) x + 5\right) \left(x + 3\right)$

$= a {x}^{3} + b {x}^{2} + \left(3 b - 9 a + 5\right) x + 15$

[2] $a {x}^{3} + b {x}^{2} - 28 x + 15 + 60$

$= \left(a {x}^{2} + \left(b + 3 a\right) x - 25\right) \left(x - 3\right)$

$= a {x}^{3} + b {x}^{2} - \left(3 b + 9 a + 25\right) x + 75$

Then looking at the coefficient of $x$ we get:

[1a] $3 b - 9 a + 5 = - 28$

[2a] $3 b + 9 a + 25 = 28$

Add [1a] and [2a] to get:

$6 b + 30 = 0$

So $b = - 5$

Subtract [1a] from [2a] to get:

$18 a + 20 = 56$

So $a = 2$

From [1] we then have:

$f \left(x\right) = \left(2 {x}^{2} - 11 x + 5\right) \left(x + 3\right)$

$= \left(2 x - 1\right) \left(x - 5\right) \left(x + 3\right)$

So the roots of $f \left(x\right) = 0$ are $x = \frac{1}{2}$, $x = 5$ and $x = - 3$