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- #1

- Feb 14, 2012

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Factor the expression

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

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

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,844

Factor the expression

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

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

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- #2

- Mar 5, 2012

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Since all combinations are present, factorization can likely be done asFactor the expression

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

$$(Aa+Bb+Cc+Dd)(\alpha a + \beta b + \gamma c + \delta d)$$

$$

\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:

$$(3a+5b-6c-15d)(10a+6b-5c-2d) \qquad \blacksquare$$

- Jan 26, 2012

- 183

$30a^2+68ab+30b^2$ 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.

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- #4

- Feb 14, 2012

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Thanks for participating to this challenge problem and my solution is quite similar to

Best regards,

anemone

- May 13, 2013

- 386

Where is the complete solution to this problem?

Anyone?

Anyone?

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