Could translate/explain this for me?

Question

I have this problem: $$ 10x^2 - 7x - 12 = 0 $$ And apparently the method to factoring it is to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$. However, I can barely understand what that even means, and as far as I'm concerned it seems impossible with the expression given. Could someone possibly just walk me through this method?

Answer 1

to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.

Take the smallest, innermost noun phrases first:

to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.

Identify those pieces in the given polynomial:

$$ \underbrace{10}_{\text{coefficient of $x^2$}}x^2 + \overbrace{(-7)}^{\text{coefficient of $x$}}x + \underbrace{(-12)}_{\text{constant term}} $$

Plug those in to the question, to simplify it:

to find two numbers whose product is the same as the product of $10$ and $-12$, and whose sum is the same as $-7$.

"The product of $10$ and $-12$" is $-120$. Let's plug that in.

to find two numbers whose product is the same as $-120$, and whose sum is the same as $-7$.

Now the words "the same as" seem pointless, so let's get rid of them:

to find two numbers whose product is $-120$, and whose sum is $-7$.

An obvious way to proceed here is just to list all the pairs of numbers whose product is $-120$, and to see which of them have sum $-7$. (We can list them all the ones where both numbers are integers, at least... hopefully that'll be enough.) Now, $-120$ has lots of factors, so there's lots of possibilities to consider. Any way of splitting up the factors on the RHS of $$ -120 = -1\cdot 2\cdot 2 \cdot 2 \cdot 3\cdot 5 $$ into two groups will make a factorization of $-120$ which might work. For example, we might try $$ -1\cdot 2\cdot 3 \qquad\text{and}\qquad 2\cdot 2\cdot 5 $$ Those are $-6$ and $20$; their product is $-120$, but their sum is $14$, not $-7$ as we wished.

By playing around for a while, or by systematically checking every possibility, you'll eventually find that $$ 2\cdot 2\cdot 2 \qquad\text{and}\qquad -1\cdot 3\cdot 5 $$ that is, $8$ and $-15$, have product $-120$ and sum $-7$.

The rest of the method, which you didn't mention, is to break up the $x$ term using these two numbers, \begin{align*} 10x^2 - 7x - 12 &= 10x^2 - 15x + 8x - 12 \end{align*} then to take out common factors from the left two terms and from the right two terms, \begin{align*} 10x^2 - 7x - 12 &= 10x^2 - 15x + 8x - 12 \\ &= 5x(2x - 3) + 4(2x - 3) \end{align*} and then, if we did everything right, we'll find, as we did here, that there's now a common factor between the left and the right, which we take out: \begin{align*} 10x^2 - 7x - 12 &= 10x^2 - 15x + 8x - 12 \\ &= 5x(2x - 3) + 4(2x - 3) \\ &= (5x+4)(2x-3) \end{align*} Factorization complete.

Answer 2

You have,

$$10x^{2}-7x-12=0$$

To factor this, I would use the quadratic formula,

that is,

$$x=\frac{7 \pm \sqrt{49-4(10)(-12)}}{20}$$

which solving leads to

$$x= \frac{-4}{5}$$ and $$x= \frac{3}{2}$$

to put this in the factored form now is easy as we can just write it as ,

$$(2x-3)(5x+4)$$ (notice the roots don't change, but I multiplied to get whole numbers)

You are also asking about a certain method.

If we rewrite your quadratic in an equivalent form

ie,

$$x^2-\frac{7x}{10}-\frac{6}{5}=0$$

Then we can use the method that is to find the two numbers that multiply to -6/5 and add to -7/10.

Do you notice anything similar from our numbers?

Answer 3

The answer is: $(2x-3)(5x+4)$. You need to do some guessing that comes with experience. In this case, you assume whole numbers for the coefficients and look at factoring - $10 = 5\cdot2$ and $12 = 3\cdot4$, and you use this a as a first attempt.

Answer 4

To answer your question: Your method may be referring to Vieta's formulas.

First, if it is possible to factor the quadratic, there always exist such (real) numbers $x_1,x_2$:

$$ax^2+bx+c=a(x-x_1)(x-x_2),\ \ a\neq 0$$

Finding the $x_1,x_2$ is your task when you are asked to factor. Then (use distributive property):

$$a(x-x_1)(x-x_2)=ax^2+x(-ax_1-ax_2)+ax_1x_2$$

$$-ax_1-ax_2=b,\ \ \ \ \ \ ax_1x_2=c$$

$$x_1+x_2=-\frac{b}{a},\ \ \ \ \ \ x_1x_2=\frac{c}{a}$$

In your case, $10x^2-7x-12=10(x-x_1)(x-x_2)$ with $$x_1+x_2=-\frac{-7}{10},\ \ \ \ \ \ x_1x_2=\frac{-12}{10}$$

$$x_1+x_2=0.7,\ \ \ \ \ \ x_1x_2=-1.2$$

$x_1=1.5,\, x_2=-0.8$ works by guessing (you can solve the system of course, but at this point you would use quadratic formula instead and it would miss the point of the method).

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In this case the method didn't work well. It can however work quite well when $a=1$ and the coefficients are integers.

E.g., see $x^2-4x-12=(x-x_1)(x-x_2)$. Then $$x_1+x_2=4,\ \ \ \ \ \ x_1x_2=-12$$

Guessing and checking can fairly easily get you $\{x_1,x_2\}=\{6, -2\}$, and so $$x^2-4x-12=(x-6)(x+2)$$

Answer 5

Suppose you start with the expression $$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd=px^2+qx+r$$

This helps us to see how the factoriastion we want relates to the polynomial we have. We know $p,q,r$ but we don't know $a,b,c,d$.

What we do know, though, is that if $a, b, c, d$ exist then comparing coefficients we get $abcd=pr$ and $ad+bc=q$. The two numbers $ad$ and $bc$ have sum $q$ and product $pr$.

Let's suppose we have any two numbers with $de=pr, e=\frac {pr}d$ (assuming $d\neq 0$, which is a trivial case) and $d+e=q$.

We then can write $$px^2+qx+r=px^2+(d+e)x+r=(px^2+dx)+(\frac {pr}dx+r)=x(px+d)+\frac rd(px+d)=$$$$=\left (x+\frac rd\right)(px+d)=\frac 1d(dx+r)(px+d)$$

Now if it matters that we are working with integers, note that $d\mid pr$ by construction, so $d$ can be cancelled.