find $a$ and $b$ where $x^3 - 4x^2 -3x + 18 = (x+a)(x-b)^2$

Question

I have a problem solving this as when I match the coefficients to the expanded brackets I end up with $2$ unknowns $a \& b$. So cannot substitute any values to find the other. According to the answers $a=2$ and $b=3$ but I cannot get there. Would someone be able to provide a worked through example so I can figure this out. Thanks.

Answer 1

$x^3 - 4x^2 -3x + 18 = (x+a)(x-b)^2 = (x+a)(x^2-2bx+b^2)=x^3+(a-2b)x^2+(b^2-2ab)x+ab^2$

$a-2b=-4$, $a=2b-4$

$b^2-2ab=-3$

$ab^2=18$

$b^2-2ab= b^2 - 2b(2b-4)=-3b^2+8b=-3$

$3b^2-8b-3=(3b+1)(b-3)=0$, $b=\frac{-1}{3}, a=\frac{-14}{3}$(NA) or $b=3,a=2$

Answer 2

your conditions say that there is a nontrivial common factor in $x^3 - 4x^2 -3x + 18 $ and its derivative $3x^2 -8x -3 $ The next to last line below is a Bezout type formula...

We can investigate that using the Euclidean algorithm

$$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) $$

$$ \left( 3 x^{2} - 8 x - 3 \right) $$

$$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) = \left( 3 x^{2} - 8 x - 3 \right) \cdot \color{magenta}{ \left( \frac{ 3 x - 4 }{ 9 } \right) } + \left( \frac{ - 50 x + 150 }{ 9 } \right) $$ $$ \left( 3 x^{2} - 8 x - 3 \right) = \left( \frac{ - 50 x + 150 }{ 9 } \right) \cdot \color{magenta}{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 3 x - 4 }{ 9 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x - 4 }{ 9 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 9 x^{2} + 9 x + 54 }{ 50 } \right) }{ \left( \frac{ - 27 x - 9 }{ 50 } \right) } $$ $$ \left( x^{2} - x - 6 \right) \left( \frac{ 9}{50 } \right) - \left( 3 x + 1 \right) \left( \frac{ 3 x - 4 }{ 50 } \right) = \left( -1 \right) $$ $$ \mbox{confirming GCD} = \color{blue}{ \left( x - 3 \right) } $$ $$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) = \left( x^{2} - x - 6 \right) \cdot \color{blue}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \left( 3 x^{2} - 8 x - 3 \right) = \left( 3 x + 1 \right) \cdot \color{blue}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \left( x^{3} - 4 x^{2} - 3 x + 18 \right) \left( \frac{ 9}{50 } \right) - \left( 3 x^{2} - 8 x - 3 \right) \left( \frac{ 3 x - 4 }{ 50 } \right) = \color{blue}{ \left( - x + 3 \right) } $$ $$ \mbox{GCD} = \color{blue}{ \left( x - 3 \right) } $$

The conclusion is that $(x-3)^2$ divides $x^{3} - 4 x^{2} - 3 x + 18.$ So, divide the original by $x^2 - 6 x + 9,$ the quotient will be some $x +a$