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How to factor $$25x^2 +5xy -6y^2$$

I tried with $5x(5x+y)-6y^2$. I'm stuck here.

I can't continue.

Pp..
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user155971
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  • What are factors of -6? – Joffan Jan 22 '15 at 15:45
  • use the quadratic formula with $a = 25, b = 5y, c = -6y^2$ to find $x.$ – abel Jan 22 '15 at 15:48
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    @Joffan, that's almost useless. We need to find the factors of $ac = -150$. – The Chaz 2.0 Jan 22 '15 at 15:52
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    Here's the general method: http://www.adamlchan.com/math/topics/ac_factoring/index.php – The Chaz 2.0 Jan 22 '15 at 15:54
  • @TheChaz2.0 the factors of -6 will be needed at some stage, and arguably the division of 25 into $5\times 5$ effectively means that it is the only step left. – Joffan Jan 22 '15 at 15:58
  • $ax^2+bx+c=a(x-x_1)(x-x_2), \forall a\neq 0, b,c,x\in\mathbb R, x_1,x_2$ are the roots of the polynomial. Thus we only need to find the roots. There are two ways you could do that: consider $25(x^2)+5y(x)-6y^2, \Delta=\cdots$. Or you can divide both sides of $5x^2+5xy-6y^2=0$ by $y^2$ and consider it as a quadratic equation: $5\left(\frac{x}{y}\right)^2+5\left(\frac{x}{y}\right)-6=0$. – user26486 Jan 22 '15 at 16:36

8 Answers8

11

Forget about the $y$ for a second.

If you want to factor $25x^2 + 5x - 6$, look for two numbers whose product is $25 \cdot -6 = -150$ and whose sum is $5$. It takes just a moment to see that the two numbers are $15$ and $-10$. Then: \begin{align*} 25 x^2 + 5x - 6 &= 25 x^2 + (15x - 10x) - 6 \\ &= (25 x^2 + 15x) - (10x + 6) \\ &= 5x(5x + 3) - 2(5x + 3) \\& = (5x-2)(5x+3).\end{align*}

With the $y$ included the process is nearly identical: \begin{align*} 25 x^2 + 5xy - 6 y^2&= 25 x^2 + (15xy - 10xy) - 6y^2 \\ &= (25 x^2 + 15xy) - (10xy + 6y^2) \\ &= 5x(5x + 3y) - 2y(5x + 3y) \\& = (5x-2y)(5x+3y).\end{align*}

Umberto P.
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    Imagine that - someone who actually knows how to teach! – The Chaz 2.0 Jan 22 '15 at 16:11
  • @TheChaz2.0 Your arbitrary pedantry amuses me. You criticized my answer for not addressing a bunch of minutia; in the same vein, you could criticize this answer (one that many people have obviously found helpful)--why do we have to look for two numbers whose product is $25\cdot -6 = -150$ and whose sum is $5$? Your fastidiousness concerned with addressing such niceties indubitably betrays your own pedagogical hubris. – Daniel W. Farlow Jan 22 '15 at 16:57
  • This is how I would do it. – JP McCarthy Jan 23 '15 at 15:11
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Hint: Factor $25x^2+5x-6$ first.

user2345215
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The trick here is to manipulate the expression $25x^2+5xy-6y^2$. Try the following: \begin{align} 25x^2+5xy-6y^2 &= 25x^2+15xy-10xy-6y^2\tag{manipulate}\\[0.5em] &= 5x(5x+3y)-2y(5x+3y)\tag{factoring}\\[0.5em] &= (5x-2y)(5x+3y)\tag{group} \end{align} Is this clear now?

Daniel W. Farlow
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    It's not clear to me how you divined that $5 = 15 - 10$ (and not some other sum) – The Chaz 2.0 Jan 22 '15 at 16:00
  • What do you mean "divined"? Knowing how to appropriately add and subtract things in a formula without changing the meaning is pretty common I think. In this case, what's "most appropriate" is $5xy=15xy-10xy$. Obviously I could write a host of other equivalent sums, but what use would they have in the context of the OP's question? – Daniel W. Farlow Jan 22 '15 at 16:02
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    Yes, but how did you "choose" $15-10$? Why not $9-4$ or $1+4$? To a student, your choice just looks like a guess, whence "divine". – The Chaz 2.0 Jan 22 '15 at 16:04
  • Call it intuition if you want. How do we really know that $1+1=2$? Maybe ask Russell and Whitehead. Your objection is not worth discussing here I don't think. I mean how would you explain why my choice was correct? – Daniel W. Farlow Jan 22 '15 at 16:08
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    You found the factor pair of $ac = -150$ that sums to $b= 5$. Your comparison is ludicrous, and pedagogy deplorable. – The Chaz 2.0 Jan 22 '15 at 16:10
  • You can beat anything to death if you want to (your grammar and use of commas are deplorable by the way). If what you think of sound pedagogy is belaboring a trivial point, then have at it. Rest easy. – Daniel W. Farlow Jan 22 '15 at 16:13
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    @TheChaz2.0 Read your auto bio on your profile--you may want to consider updating the link you provide to your funded board game (the link is broken). – Daniel W. Farlow Jan 22 '15 at 16:17
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Instead of finding all of the factors of -150, lets do this a bit more intelligently. Notice that 25 and 5 are both multiple of 5. This gives us the opportunity to substitute $t=5x$ $$25x^2+5xy-6y^2=(5x)^2+(5x)y-6y^2=t^2+ty-6y^2=\dots$$ After factoring, just substitute the $5x$ back in.

John Joy
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Or just use $abc$ formula, with $a = 25$, $b = 5y$ and $c = -6y^2$, and find that terms disappear like snow before the sun:

$$ \frac{-5y ^+_- \sqrt{625y^2} }{2 \times 25}$$

So $5x = -3y$ or $5x = 2y$, etc.

Pieter21
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let me expand on my comment of using quadratic formula to factor an expression like this. of course, you need to have the quadratic formula that $x = \dfrac{-b + \pm \sqrt{b^2 - 4ac}}{2a}$ are the solutions of $ax^2 + bx + c = 0.$

we will look at first the equation $25x^2 +5xy - 6y^2 = 0$ later on we will see how we can use the solutions to factor the expression. matching $25x^2 +5xy - 6y^2 = 0$ with $ax^2 + bx + c = 0$ we can identify $a = 25, b = 5y, c = -6y^2.$ first the discriminant $b^2 - 4ac = (5y)^2 - 4*25*(-6y^2) = 625y^2 = (25y)^2$

using the formula $x = \dfrac{-5y \pm 25y}{50} = -\dfrac{3y}{5}, \dfrac{2y}{5}$

we can turn this into two equations into the factors $k(x+\frac{3y}{5})(x - \frac{2y}{5}) = 25x^2 +5xy - 6y^2 = 0 $ matching the coeffcient of $x^2$ fixes $k = 25.$

what do we have now is $$25(x+\frac{3y}{5})(x - \frac{2y}{5})= (5x + 3y)(5x - 2y)= 25x^2 +5xy - 6y^2 = 0 $$

from the last equalities drop leftmost and the rightmost expressions and you have factored $25x^2 +5xy - 6y^2.$

abel
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Let me add some explanations on Umberto P.'s answer, which is the same as that given in http://www.adamlchan.com/math/topics/ac_factoring/index.php

cited by The Chaz 2.o.

Assume that $$Ax^2+Bxy+Cy^2=(Px+Qy)(Rx+Sy)$$

Then $$Ax^2+Bxy+Cy^2=(PR)x^2+(PS+QR)xy+(QS)y^2$$

Let's first find $(PS)$ and $(QR)$.

Step 1: Calculate $AC$.

Then observe that $$(PS)(QR)=AC$$ $$PS+QR=B$$

Step 2: Find two numbers whose product is $AC$ and whose sum is $B$.

After obtaining $PS$ and $QR$,

Step 3: Rewrite the original equation as $(PR)x^2+(PS)xy+(QR)xy+(QS)y^2$

Step 4: Grouping the terms and factorize them $$[(PR)x^2+(PS)xy]+[(QR)xy+(QS)y^2]=Px(Rx+Sy)+Qy(Rx+Sy)$$

Step 5: Applying distributive law $$Px(Rx+Sy)+Qy(Rx+Sy)=(Px+Qy)(Rx+Sy)$$

Applying to this problem:

Step 1: $25\times(-6)=-150$

Step 2: Find two numbers whose product is -150 and whose sum is 5 --> 15 and -10

Step 3: Write it as $25x^2+15xy-10xy-6y^2$

Step 4: $(25x^2+15xy)+(-10xy-6y^2)=5x(5x+3)-2y(5x+3y)$

Step 5: $5x(5x+3)-2y(5x+3y)=(5x-2y)(5x+3y)$

velut luna
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The answer is $(5x-2y)(5x+3y)$.

velut luna
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