How Can I Factorise Equations with 2 Brackets?

[Gringo123]

Can anyone help me with the following 2 factorising problems?

x squared - 7 squared + 12

and
6x squared + 5x - 4

I know that the answers are:
(x-3) (x-4)
and
(3x + 4) (2x - 1)

I guess that in the first one, the solution has something to do with the fact that 3 & 4 add together to make 7 amd multiply to make 21?

 

[Char. Limit]

If by "multiply to make 21" you mean "multiply to make 12", yes. Simple typo, no problem.

Of course, you are dealing with -3 and -4 here, the middle term (-7x, not -7 squared) is the sum of these two and the last term (12 here) is the product of these two.

Of course, once you start using x coefficients it gets harder (3x and 2x) to see the solutions intuitively.

 

[Mark44]

Can anyone help me with the following 2 factorising problems?

x squared - 7 squared + 12

From your answer, the expression you want to factor (as we say it here in the US) is x² - 7x + 12.

The idea is to factor this trinomial into a product of two binomial factors.
x² - 7x + 12 = (x + ?)(x + ?)
What you need to do is find two factors of 12 that add up to -7

You really should learn how to write the symbols. You can write exponents by using the X² button in the menu bar above the input window. If that menu bar doesn't appear, click the Go Advanced button just below the input window.

Also (and again) these -- () -- are parentheses, not brackets. These -- [] -- are brackets.

and
6x squared + 5x - 4

You want two binomial factors that multiply to the above. Different possibilities are:
(6x + ?)(x + ?)
(3x + ?)(x + ?)

Where I have the ? characters, you need to have factors of -4: -4 * 1, 4 * -1, -2 * 2, 2 * -2. One of these combinations combined with one of the two binomial pairs of factors above might be the one that works. Be advised, though, that it's not possible to factor all trinomials into binomial factors with integer coefficients.

I know that the answers are:
(x-3) (x-4)
and
(3x + 4) (2x - 1)

I guess that in the first one, the solution has something to do with the fact that 3 & 4 add together to make 7 amd multiply to make 21?

 

[Sirsh]

Can anyone help me with the following 2 factorising problems?

(x²-7x²+21)
and
(6x²+5x-4)x(x+7) - 21

x(6x+5) - 4

 

[Mentallic]

Can anyone help me with the following 2 factorising problems?

(x²-7x²+21)
and
(6x²+5x-4)


x(x+7) - 21

x(6x+5) - 4

You should make your own thread.

Gringo123, maybe this will shed some light on why you need to do what you're doing.
(by the way, this only works for the first problem where the coefficient of x² is 1)

We have (x+a)(x+b) where a and b are some numbers. If we expand this we get: x²+ax+bx+ab=x²+(a+b)x+ab

Now, we have the problem x²-7x+12 to factorize.

So, $x^2+(a+b)x+ab\equiv x^2-7x+12$ and equating coefficients, this means that

a+b=-7
ab=12

If you can solve these simultaneously, you'll have your answers of a and b.

 

[Sirsh]

You should make your own thread.

Sorry I was just giving factorisations of those two expressions forgot that i left the question in the post

 

[Mentallic]

Aha. Sorry I didn't notice it was the same question as the original question the OP gave.
But then, your answers aren't completely factorized.

 

[Sirsh]

I believe my answers are more easiliy worked with rather then the double brackets, just showing there are two different ways of doing it.

 

[Mentallic]

I disagree.
In what circumstances would your form of factorizing be easier to work with? You can't find the roots of the function y=x(x+7)-21 easily. Also it doesn't make things much easier to graph.

However, completing the square is another way of factorizing that is useful when drawing the graph.

e.g. $x^2-7x+12=(x-7/2)^2-1/4$ tells us the quadratic's minimum point is at (7/2,-1/4).

 

[Char. Limit]

I find that an easier way to calculate the min/max is to find the two roots. The min/max is almost always halfway between them.

Example, $$x^2-1 = (x+1)(x-1)$$.

The root is halfway between -1 and +1, or zero. Look on the graph and this is the case.

Of course, this wouldn't work for a graph such as $$x^2+1 = (x+i)(x-i)$$... wait, it does. This can't be the case for every quadratic with no real roots, however. Or maybe it can. I just know that it does work for most every quadratic with one or two real roots.

 

[Mentallic]

Of course, this wouldn't work for a graph such as $$x^2+1 = (x+i)(x-i)$$... wait, it does. This can't be the case for every quadratic with no real roots, however. Or maybe it can. I just know that it does work for most every quadratic with one or two real roots.

I just realized it works as well!

And yeah, for every real quadratic it can be done since the requirements for a real quadratic are that if one root is complex, then the other is the complex conjugate.

e.g. $$x^2-2x+2=(x-1)^2+1=(x-(1+i))(x-(1-i))$$

The thing is though that you must first find the average of the roots to find the axis of symmetry. After this you need to plug it back into the formula to get the y-value. Completing the square does both of these steps in one... so the work load is less overall.

Also, notice that to find the roots of the above example you need to use the quadratic formula, which is another hassle that's unnecessary.

 

[Sirsh]

I disagree.
In what circumstances would your form of factorizing be easier to work with? You can't find the roots of the function y=x(x+7)-21 easily. Also it doesn't make things much easier to graph.

Yes, I've got to agree with you. completely forgot about roots of the functions. (: