How to factor $3x^2-4q^2-3p^2+4qx-8pq$?

Question

I only know the usual techniques for factoring. But I came across this problem:

$$3x^2-4q^2-3p^2+4qx-8pq$$

I couldn't factorise this but thanks to the solution booklet, I found the answer to be:

$$(x+p+2q)(3x-3p-2q)$$

Can somebody help me to factorise this? Thanking you in advance!

Answer 1

Let $$3x^2-4q^2-3p^2+4qx-8pq=(ax+bp+cq)(dx + ep + fq)$$

Opening R.H.S. and comparing factors, we get that ;

$$ad= 3 , be=-3, cf=-4, ae+bd = 0,ce+fb=-8,cd+fa=-4$$

Solving these equations(which is an exercise for you), we get ;

$(a,b,c,d,e,f)= (1,1,2,3,-3,-2)$ and $(3,-3,-2,1,1,2)$

Hence, $$3x^2-4q^2-3p^2+4qx-8pq = (x+p+2q)(3x-3p-2q)$$

Answer 2

A quadratic form in three variables factors (perhaps with complex coefficients) if and only if the following procedure says so. In brief, the determinant of the Hessian matrix must be zero for there to be any possible factoring.

The relation $Q^T D Q = H$ shows that your quadratic form is $$ 3 \left(x + \frac{2q}{3} \right)^2 - \frac{16}{3} \left(q + \frac{3p}{4} \right)^2 = \frac{1}{3} \left( 9 \left(x + \frac{2q}{3} \right)^2 -16 \left(q + \frac{3p}{4} \right)^2 \right) $$ $$ = \frac{1}{3} \left( \left(3x + 2q \right)^2 - \left(4q + 3p \right)^2 \right) = \frac{1}{3} \left(3x + 6q + 3p \right) \left(3x - 2q - 3p \right) $$ $$= \left(x + 2q + p \right) \left(3x - 2q - 3p \right) $$ which factored as a difference of squares

$$ P^T H P = D $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 2 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 2 } & - \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 2 & 0 \\ 2 & - 4 & - 4 \\ 0 & - 4 & - 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 2 }{ 3 } & \frac{ 1 }{ 2 } \\ 0 & 1 & - \frac{ 3 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & - \frac{ 16 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$

$$ Q^T D Q = H $$

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 2 }{ 3 } & 1 & 0 \\ 0 & \frac{ 3 }{ 4 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & - \frac{ 16 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 2 }{ 3 } & 0 \\ 0 & 1 & \frac{ 3 }{ 4 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 2 & 0 \\ 2 & - 4 & - 4 \\ 0 & - 4 & - 3 \\ \end{array} \right) $$

Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices

Answer 3

Hint: $\;3x^2 + 4qx - (4q^2 + 8pq + 3p^2)$ as a quadratic in $x$ has the reduced discriminant:

$$ \frac{1}{4}\Delta = 4q^2 + 3(4q^2 + 8pq + 3p^2) = 16 q^2 + 24 pq + 9p^2 = (4q + 3p)^2 $$

Answer 4

If we treat the effort at factorization as equivalent to solving the quadratic equation $ \ 3x^2 + 4qx - (3p^2 + 8pq + 4q^2) \ = \ 0 \ \ , $ "completing the square" produces $$ 3 · \left[ \ x^2 \ + \ \frac{4q}{3}·x \ \right] \ - \ ( \ 3p^2 \ + \ 8pq \ + \ 4q^2 \ ) $$ $$ = \ \ 3 · \left[ \ x^2 \ + \ \frac{4q}{3}·x \ + \ \left(\frac{2q}{3} \right)^2 \ \right] \ - \ ( \ 3p^2 \ + \ 8pq \ + \ 4q^2 \ ) \ - \ 3·\frac{4q^2}{9} $$ $$ = \ \ 3 · \left( \ x \ + \ \frac{2q}{3} \right)^2 \ - \ \left( \ 3p^2 \ + \ 8pq \ + \ \frac{16}{3}q^2 \ \right) $$ $$ = \ \ 3 · \left[ \ \left( \ x \ + \ \frac{2q}{3} \right)^2 \ - \ \left( \ p^2 \ + \ \frac{8}{3}pq \ + \ \frac{16}{9}q^2 \ \right) \ \right] \ \ = \ \ 0 \ \ . $$

It is now a bit more evident that the second term in parentheses is a "binomial-square" and that the equation is solved by

$$ \left( \ x \ + \ \frac{2q}{3} \right)^2 \ - \ \left( \ p \ + \ \frac{4}{3}q \ \right)^2 \ \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ -\frac{2q}{3} \ \pm \ \left( \ p \ + \ \frac{4}{3}q \ \right) $$ $$ \Rightarrow \ \ x \ \ = \ \ p \ + \ \frac{4}{3}q \ - \ \frac{2}{3}q \ \ = \ \ p \ + \ \frac{2}{3}q \ \ \ , \ \ \ x \ \ = \ \ -p \ - \ \frac{4}{3}q \ - \ \frac{2}{3}q \ \ = \ \ -p \ - \ 2q \ \ . $$

These solutions and the Factor Theorem tells us that, with the leading coefficient of the quadratic polynomial being $ \ 3 \ \ , \ $ the factorization is $$ 3x^2 \ + \ 4qx \ - \ (3p^2 + 8pq + 4q^2) \ \ = \ \ 3 \ · \ (x \ + \ p \ + \ 2q) \ · \ \left(x \ - \ p \ - \ \frac{2}{3}q \right) $$ $$ \text{or} \ \ \ (x \ + \ p \ + \ 2q) \ · \ (3x \ - \ 3p \ - \ 2q ) \ \ . $$

Answer 5

Let $x = q$. Then we have $3q^2-4q^2-3p^2+4q^2-8pq = 3q^2-8pq-3p^2 = (3q + p)(-3p + q)$.

Now let $x = p$. We get $3p^2 - 4q^2 - 3p^2 + 4pq - 8pq = -4q^2 - 4pq = -4q(q + p)$.

Now finally let $p = q$. We get $3x^2 - 4q^2 - 3q^2 + 4qx - 8q^2 = 3x^2 + 4qx - 15q^2 =(3x - 5q)(x + 3q)$.

The union of these factorisations contains all the variables $x, p$, and $q$, so the original factorisation must contain all of these too.

Now we need to know which brackets in the reduced factorisations correspond with each other. Observe that substituting any variable for another does not affect the sum of the coefficients. For example, if we replace $-5q$ with $-3q - 2p$, the sum of the coefficients in that bracket is unchanged. So it is pretty clear that $(3x - 5q)$ corresponds to $(-3p + q)$, and $(x + 3q)$ corresponds to $(3q + p)$.

Let us try to find the full form of the bracket $3x - 5q$. If we choose $3x - 6q + p$ for example, then this equals $3x - 5q$ when $p = q$, but not $-3p + q$ when $x = q$. It thus occurs that our coefficient of $p$ must match $-3p$ as $p$ is unaffected when $x = q$. Thus one bracket must have $-3p$ (when $x = q$), $3x$ (when $p = q$), and $-2q$ ($-3-2 = -5$ when $p = q$). Hence that bracket must be $(-3p - 2q + 3x)$.

Similarly, in the other bracket we must have $p$, $x$, and a $(3 - 1)q = 2q$.

Thus our complete factorisation is $(-3p - 2q + 3x)(p + 2q + x)$, which can be verified by expanding the brackets and comparing terms.