Math CentralQuandaries & Queries


Question from pedro, a student:

factor 6x2-19x+3

Hi Pedro,

You want to write $6 x^2 - 19 x + 3 = (ax + b)(cx + d)$ for some numbers $a, b, c \mbox{ and } d.$ If you imagine multiplying the right side to get the left side you have $a \times c = 6, \mbox{ and } b \times d = 3.$ From the second of these equations you must have $b$ and $d$ either 1 and 3 or -1 and -3.

If a, b, c and d are all positive then expanding $(ax + b)(cx + d)$ will yield all positive terms but the original expression has a negative middle term, $-19 x$ and hence I am going to try $b = -1$ and $d = -3.$ Thus I have

\[6 x^2 - 19 x + 3 = (ax -1)(cx -3).\]

But I know that $a \times c = 6.$ What are the possibilities for $a$ and $c?$ Try them and expand the right side to see which choice of $a$ and $c$ give a middle term of $-19 x.$


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