How do you factor 49x^2 - 14x - 349x2−14x−3?

1 Answer
Alan P.
Mar 28, 2015

Since 49x^2-14x-349x2−14x−3 is a quadratic of the form ax^2+bx+cax2+bx+c
we could use the formula (which you really should memorize)
(-b +- sqrt(b^2-4ac))/(2a)−b±√b2−4ac2a

But first let's see if there is a simpler method under the assumption that we are dealing solely with integers (not necessarily true, but the work is easier if it is true).

We are hoping to find something of the form
(px+q)(rx+s) = 49x^2-14x-3(px+q)(rx+s)=49x2−14x−3

If all values are integer there are only 3 possibilities for (p,r)(p,r)
namely (1,49)(1,49), (49,1)(49,1) and {7,7}{7,7}

The (1,49)(1,49) and (49,1)(49,1) pairs don't look too likely given the relatively small coefficient of xx (we'll return to them later if necessary), so let's try (p,r) = (7,7)(p,r)=(7,7)

We are hoping to find integer values qq and ss such that
(7x+q)(7x+s) = 49x^2-14x-3(7x+q)(7x+s)=49x2−14x−3

For this to work
7q+7s = 147q+7s=14 which implies q+s = 2q+s=2
and
qs= -3qs=−3

Again with limited possibilities for integer values of qq and ss where qs = -3qs=−3

We can quickly deduce that the required factorization is
(7x-3)(7x+1)(7x−3)(7x+1)

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