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Problem Of The Week #388 Oct 17th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to the following members for their correct answer!(Cool)

1. kaliprasad
2. Greg

Solution from kaliprasad :
$(x+1)(x+2)(x+3)(x+6) - 3x^2$
= $(x+1)(x+6)(x+2)(x+6)- 3x^2$
= $(x^2 + 7x + 6)(x^2 + 5x + 6) - 3x^2$
= $((x^2 + 6x+ 6) + x)((x^2 + 6x + 6) -x) - 3x^2$
= $(x^2+ 6x + 6)^2 - x^2 - 3x^2$
= $(x^2 + 6x+6)^2 - 4x^2$
= $(x^2 + 6x + 6)- (2x)^2$
= $(x^2 + 8x + 6) (x^2 + 4x+ 6)$ (these 2 cannot be factored further)


Alternate solution from Greg :
\(\displaystyle P(x)=(x+1)(x+2)(x+3)(x+6)-3x^2=x^4+12x^3+44x^2+72x+36\)

By inspecting $P(x)$ in the context of the Rational Roots theorem we see that $P(x)$ has no linear factors so, if $P(x)$ factors, it must be a biquadratic.

In general,

$$(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$$

so we have

$$a+c=12\quad ac+b+d=44\quad ad+bc=72\quad bd=36$$

By inspection,

$$a=8\quad b=6\quad c=4\quad d=6$$

and $P(x)=(x^2+8x+6)(x^2+4x+6)$ as required.
 
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