How can I find the common quadratic factor of multiple polynomials?

Question

$x^4-13x^2+36$,
$2x^3+3x^2-11x-6$,
$3x^3+x^2-20x+12$
Find a common quadratic factor of these expressions. Help?

Answer 1

Letting these polynomials be $p_1(x),p_2(x),p_3(x)$, then any common divisor of $p_2$ and $p_3$ must be a divisor of:

$$3p_2(x)-2p_3(x)=7x^2+7x-42=7(x^2+x-6)$$

Does $x^2+x-6=(x+3)(x-2)$ divide $p_1$, $p_2$, and $p_3$?

Answer 2

First and second: $$ \left( x^{4} - 13 x^{2} + 36 \right) $$

$$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) $$

$$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) \left( \frac{ 4}{21 } \right) - \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) \left( \frac{ 2 x - 3 }{ 21 } \right) = \left( - x^{2} - x + 6 \right) $$

First and third: $$ \left( x^{4} - 13 x^{2} + 36 \right) $$

$$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) \left( \frac{ 9}{56 } \right) - \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \left( \frac{ 3 x - 1 }{ 56 } \right) = \left( - x^{2} - x + 6 \right) $$ Second and third: $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) $$

$$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) \left( \frac{ 3}{7 } \right) - \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \left( \frac{ 2}{7 } \right) = \left( x^{2} + x - 6 \right) $$

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more detail on the Euclidean algorithm

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$$ \left( x^{4} - 13 x^{2} + 36 \right) $$

$$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) $$

$$ \left( x^{4} - 13 x^{2} + 36 \right) = \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } + \left( \frac{ - 21 x^{2} - 21 x + 126 }{ 4 } \right) $$ $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) = \left( \frac{ - 21 x^{2} - 21 x + 126 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 21 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 2 x - 3 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 3 }{ 4 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 21 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 4 x^{2} + 4 x + 24 }{ 21 } \right) }{ \left( \frac{ - 8 x - 4 }{ 21 } \right) } $$ $$ \left( x^{2} - x - 6 \right) \left( \frac{ 4}{21 } \right) - \left( 2 x + 1 \right) \left( \frac{ 2 x - 3 }{ 21 } \right) = \left( -1 \right) $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) = \left( x^{2} - x - 6 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) = \left( 2 x + 1 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) \left( \frac{ 4}{21 } \right) - \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) \left( \frac{ 2 x - 3 }{ 21 } \right) = \left( - x^{2} - x + 6 \right) $$

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$$ \left( x^{4} - 13 x^{2} + 36 \right) $$

$$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) $$

$$ \left( x^{4} - 13 x^{2} + 36 \right) = \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \cdot \color{magenta}{ \left( \frac{ 3 x - 1 }{ 9 } \right) } + \left( \frac{ - 56 x^{2} - 56 x + 336 }{ 9 } \right) $$ $$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) = \left( \frac{ - 56 x^{2} - 56 x + 336 }{ 9 } \right) \cdot \color{magenta}{ \left( \frac{ - 27 x + 18 }{ 56 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 3 x - 1 }{ 9 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 3 x - 1 }{ 9 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - 27 x + 18 }{ 56 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 9 x^{2} + 9 x + 54 }{ 56 } \right) }{ \left( \frac{ - 27 x + 18 }{ 56 } \right) } $$ $$ \left( x^{2} - x - 6 \right) \left( \frac{ 9}{56 } \right) - \left( 3 x - 2 \right) \left( \frac{ 3 x - 1 }{ 56 } \right) = \left( -1 \right) $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) = \left( x^{2} - x - 6 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) = \left( 3 x - 2 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( x^{4} - 13 x^{2} + 36 \right) \left( \frac{ 9}{56 } \right) - \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \left( \frac{ 3 x - 1 }{ 56 } \right) = \left( - x^{2} - x + 6 \right) $$

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$$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) $$

$$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) $$

$$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) = \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \cdot \color{magenta}{ \left( \frac{ 2}{3 } \right) } + \left( \frac{ 7 x^{2} + 7 x - 42 }{ 3 } \right) $$ $$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) = \left( \frac{ 7 x^{2} + 7 x - 42 }{ 3 } \right) \cdot \color{magenta}{ \left( \frac{ 9 x - 6 }{ 7 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 2}{3 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2}{3 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 9 x - 6 }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 6 x + 3 }{ 7 } \right) }{ \left( \frac{ 9 x - 6 }{ 7 } \right) } $$ $$ \left( 2 x + 1 \right) \left( \frac{ 3}{7 } \right) - \left( 3 x - 2 \right) \left( \frac{ 2}{7 } \right) = \left( 1 \right) $$ $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) = \left( 2 x + 1 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \left( 3 x^{3} + x^{2} - 20 x + 12 \right) = \left( 3 x - 2 \right) \cdot \color{magenta}{ \left( x^{2} + x - 6 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x^{2} + x - 6 \right) } $$ $$ \left( 2 x^{3} + 3 x^{2} - 11 x - 6 \right) \left( \frac{ 3}{7 } \right) - \left( 3 x^{3} + x^{2} - 20 x + 12 \right) \left( \frac{ 2}{7 } \right) = \left( x^{2} + x - 6 \right) $$ jagy@phobeusjunior:~$

Answer 3

Expounding on what John Doe said in the comments:

Turns out we can factor the first polynomial easily as $$x^4 - 13x^2 + 36 = (x^2-4)(x^2-9)=(x+2)(x-2)(x+3)(x-3)$$

Any quadratic factor of the first polynomial is a product of two of these linear factors. Thus the common quadratic factor has to be a product of two of these linear factors.

These two linear factors will also be factors of the other two polynomials. To check whether each linear factor of the form $(x-a)$ is a factor of a polynomial, we need to plug $a$ into the polynomial and see if we get zero. Since we get zero in both other polynomials when we plug in both $3$ and $-2$, the common quadratic factor is $(x-3)(x+2)$.