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I am trying to factorise $2x^2+5x-12$ from Stewart's Calculus Early Transcendental diagnostic test part A (Algebra). The question is to factorise $2x^2+5x-12$ in the form of $(ax+b)(cx+d), a,b,c,d\in \mathbb{R}$. I am using the following table as I am familiar of using quadratic formula,

$2x^2$ $+5x$ $-12$
$2x$ $8x$ $-3$
$x$ $-3x$ $4$

And the answer is $(2x-3)(x+4)$.

But now, I am think about that whether completing the square method can still produce in the form of $(ax+b)(cx+d)$? Here is my working, $$2x^2+5x-12$$ $$=2(x^2+\frac{5}{2}x-6)$$ $$=2(x^2+\frac{5}{2}x+\frac{25}{16}-\frac{25}{16}-6)$$ $$=2((x+\frac{5}{4})^2-\frac{25}{16}-\frac{96}{16})$$ $$=2((x+\frac{5}{4})^2-\frac{121}{16})$$ $$=2[(\frac{1}{4}(4x+5))^2-\frac{121}{16}]$$ $$=2[(\frac{1}{16}(4x+5)^2)-\frac{121}{16}]$$ $$=\frac{1}{8}((4x+5)^2-121)$$

But then I have no where to go and I think something goes wrong in my working. :(

I know in Calculus or indeed in mathematics, there are several possible methods to find the solution and I am trying to get as many methods as possible, yeh, trying to 'getting my hands dirty'.

My tears are coming out from my eyes at the moment. T_T

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    $\begingroup$ Completing the square is actually how you find the quadratic formula, so you’re not avoiding it :) $\endgroup$ Commented Jul 6 at 14:16

2 Answers 2

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Not wrong :) Just one more step: $$(4x+5)^2-121=(4x+5)^2-11^2=(4x+5+11)\cdot(4x+5-11)$$

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  • $\begingroup$ I got it, I got it. Just managed to get $(2x-3)(x+4)$ with your help, thanks @boweitang. :) $\endgroup$ Commented Jul 7 at 4:42
  • $\begingroup$ @HectorLai You are welcome :) $\endgroup$ Commented Jul 7 at 4:45
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Given $ax^2 + bx + c$ such that the discriminant $b^2 - 4ac$ is a perfect square, take $$ \color{blue}{b^2 - 4ac = \delta^2,} $$ and $$ \color{blue}{ a_1 = \gcd \left( a, \frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1},} $$ then $$ \color{blue}{ a x^2 + b x + c = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) \right) \; } $$ in integers.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

With $2x^2+5x-12$ we get $b^2 - 4ac = 25 - 4 \cdot 2 \cdot (-12) = 25 + 96 = 121 = 11^2.$ Then $b+\delta = 5 +11 = 16,$ half of that is $8,$ and we have $a_1 = \gcd(2,8) = 2.$ So $a_2 = 1.$ Next and $2a_1 = 4 $ and the first factor is $x + 4.$ With $b-\delta = 5 -11 = -6,$ and $2 a_2 = 2,$ the second factor is $2x-3.$

Checking $(x+4)(2x-3) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Let's make up a problem.... pick $\delta = 103 $ so $\delta^2 = 10609 $ And pick $b = 47,$ so $b^2 = 2209$ and $2209 - 4ac = 10609,$ then $-4ac = 8400$ and $ac = - 2100.$ Alright, $a= 35, c = -60$ works, so we are factoring $35x^2 + 47 x - 60. \;$ We need $(b+\delta)/ 2 = 150/2 = 75.$ Then $a_1 = \gcd(35, 75) = 5.$ With $a_2 = 35/5 = 7.$ Also $(b-\delta)/ 2 = -56/2 = -28.$ Finish up $35x^2 + 47 x - 60 = (7x + 15)(5x -4)$

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  • $\begingroup$ I don't understand how you define $a_1$ and $a_2$ this way and what on earth why there is $\text{gcd}$ operator in $a_1$? $\endgroup$ Commented Jul 7 at 4:49
  • $\begingroup$ @HectorLai it is necessary to express $a=a_1 a_2$ such that all the coefficients in the factoring come out integers. If we use the gcd we don't need to be able to factor $a$ itself, we can find the gcd with the Euclidean algorithm, which is quick. $\endgroup$ Commented Jul 7 at 17:02

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