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I am making an attempt to factorise the above equation.
To do this i am expanding the already factored part of the above equation to obtain ->
$\,f(a) = a^4 - 8a^3 + 15a^2 + 4a - 20\,$
Now i do some random guesses to find that f(2) = 0, hence i know that (a-2) will divide f(a) so i perform long division to get f(a) = (a-2)( Something ) and so on... until i get - > $\,f(a) = (a - 2)^2(a + 1)(a - 5)\,$

I feel that i am approaching the problem in an incorrect and long way. Please let me know if i can get the factorised form in a cleaner and faster way.

PS: Another linked question is to factorise $\,f(x) = (x + 1)(x + 3)(x + 5)(x + 7) + 15\,$

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$1)$ Call $x=a(a-4)$ then

$$x(x-1)-20=x^2-x-20=(x+4)(x-5)$$

then

$$(a^2-4a+4)(a^2-4a-5)=(a-2)^2(a+1)(a-5)$$

$2)$ For

$$(x+1)(x+3)(x+5)(x+7)+15=[(x+3)(x+5)][(x+1)(x+7)]+15=(x^2+8x+15)(x^2+8x+7)+15$$

Call $y=x^2+8x+7$ then:

$$(y+8)y+15=y^2+8y+15=(y+3)(y+5)$$

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  • $\begingroup$ perfect! very clear $\endgroup$ – iajnr Jan 26 '17 at 17:38
  • $\begingroup$ and for f(x)=(x+1)(x+3)(x+5)(x+7)+15 i can take x^2 + 8x = t ... $\endgroup$ – iajnr Jan 26 '17 at 17:39
  • $\begingroup$ Yes that's correct. $\endgroup$ – Kanwaljit Singh Jan 26 '17 at 17:40
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Another linked question is to factorise $\,f(x) = (x + 1)(x + 3)(x + 5)(x + 7) + 15\,$

Let $z=x+4$ then:

$$ \begin{align} (z-3)(z-1)(z+1)(z+3)+15 & = (z^2-1)(z^2-9)+15=z^4-10z^2+24 \\ & = (z^2-4)(z^2-6) = (x^2+8x+12)(x^2+8x+10) \\ & = (x+2)(x+6)(x^2+8x+10) \end{align} $$

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