Factoring Algebraic Expression

[anemone]

Factor the expression

\(\displaystyle 30(a^2+b^2+c^2+d^2)+68ab-75ac-156ad-61bc-100bd+87cd\)

 

[Klaas van Aarsen]

Factor the expression

\(\displaystyle 30(a^2+b^2+c^2+d^2)+68ab-75ac-156ad-61bc-100bd+87cd\)

Since all combinations are present, factorization can likely be done as
$$(Aa+Bb+Cc+Dd)(\alpha a + \beta b + \gamma c + \delta d)$$

Spoiler

From the coefficients of the squares we can conclude that:
$$
\alpha=\frac {30} A, \beta=\frac {30} B, \gamma=\frac {30} C, \delta=\frac {30} D
$$
Let's pick $-61bc$ to evaluate.

We get:
$$\begin{array}{l}
B\gamma + C\beta=-61 \\
B \frac{30}C + C\frac{30}B=-61 \\
B=-\frac 5 6 C \vee B=-\frac 6 5 C
\end{array}$$
Since we have a free choice for what goes left and what goes right, and we also have a free choice how to divide the constant 30, we can choose:
$$B = 5 \wedge C = -6$$
Repeating for the coefficients of $ab$, $ac$, and $ad$, we find:
$$A=3 \wedge (D=-15 \vee D=-\frac 3 5)$$
Verification shows that only $D=-15$ fits, which gives indeed a solution.

In other words, the expression factorizes as:
$$(3a+5b-6c-15d)(10a+6b-5c-2d) \qquad \blacksquare$$

 

[Jester]

I might do it somewhat different. First I would group some of the terms as follows

$30a^2+68ab+30b^2$ and $30c^2+87cd+30d^2$

and factor each of these separately. This gives

$2(3a+5b)(5a+3b)$ and $3(2c+5d)(5c+2d)$.

Then let $3a+5b = u$, $5a+3b = v$, $2c+5d = p$, and $5c+2d=q$.

Solve for $a, b, c$ and $d$ and substitute into the entire expression giving

$ 2 v u + 3 q p - 6 v p - u q$

and factor this.

 

[anemone]

Hi I like Serena and Jester,

Thanks for participating to this challenge problem and my solution is quite similar to I like Serena's approach and I'll post my solution here later today.

Best regards,

anemone

 

[paulmdrdo]

Where is the complete solution to this problem?
Anyone?