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I was factorising the equation $x^2 - 3x + 10$ and realised that I can do it as $x^2 -5x + 2x + 10$, since $-5x +2x = -3x$ and $5 \cdot 2 = 10$. However, this doesn't work. As we don't get the common terms in the brackets $x(x-5) + 2(x+5)$

Is this mistake because we need $-10$ instead of $10$? That is, while checking out the multiplication of factors of $a$ and $c$, we need to make sure that it equals $ac$ along with its sign ($+$ or $-$)?

Sammy Black
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Ruchi
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2 Answers2

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You are right that this doesn't work, but this is because you have $x-5$ in one bracket and $x+5$ in the other. Recall the distributive law, shown in this image with numbers:

enter image description here

Multiplying two numbers can be thought of the area of a rectangle having the numbers as side lengths. When we have rectangles of different dimensions, there is no way to combine them to make a larger rectangle with the same width or height, unless the sides have a common factor.

The same applies for variables, so that there is no way to combine a rectangle with sides $x, x-5$ and another one with sides $2, x+5$ as in your example.

Toby Mak
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  • Thank you for your answer, but if we have another quadratic equation that could still be factorise, for example x^2 -3x+10, here 5 and 2 works. Is this because 5x2 = 10? (The signs are also correct I mean) – Ruchi Mar 01 '21 at 08:26
  • $5$ and $2$ do not work. If you try again with plus/minus $1$ and $10$, they do not work, so there are no solutions which are rational numbers. – Toby Mak Mar 01 '21 at 08:26
  • Are you sure you don't mean $x^2-3x-10$? – Toby Mak Mar 01 '21 at 08:29
  • I meant $x^2 + 3x -10 $ can be written as $ x^2 - 5x +2x -10$ and this factorises. My doubt is is it a necessary condition as well that factors of a.c ( here 5 and 2) should be equal to a.c (-10) considering the negative/positive sign or is it just the magnitude (10) ? – Ruchi Mar 01 '21 at 08:42
  • $x^2 + 3x - 10$ does not factorise. You can't get anything from $x^2 - 5x + 2x - 10$. – Toby Mak Mar 01 '21 at 08:43
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    It's + 5x - 2x sorry! Then it does factorise. Is I This because there is a condition that -5•2 = -10 or we just need the magnitude? – Ruchi Mar 01 '21 at 08:50
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It works but you can no more guess. So or the roots of $Ax^2+Bx+C=0$ $$x_{1,2}=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$$

Also see: Getting back to the quadratic equation again

Z Ahmed
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