MHB How Can You Factorize the Polynomial (x+1)(x+2)(x+3)(x+6)-3x^2?

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The polynomial to factor is (x+1)(x+2)(x+3)(x+6)-3x^2. Members kaliprasad and Greg provided correct solutions to the problem. Kaliprasad's solution was recognized as the main answer, while Greg offered an alternate method. The discussion emphasizes the importance of following the guidelines for problem-solving. Engaging with the community enhances understanding of polynomial factorization techniques.
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Here is this week's POTW:

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Factorize $(x+1)(x+2)(x+3)(x+6)-3x^2$.

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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:
$$P(x)=(x+1)(x+2)(x+3)(x+6)-3x^2=x^4+12x^3+44x^2+72x+36$$

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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