Product of numbers Pair of numbers whose product is $+7$ and whose sum is $-8$. 
Factorise $x^{2} - 8x + 7$.
I can factorise but it's just I can't find any products of $+7$ and that is a sum of 
$-8$. Any idea? Thanks guys!
Thanks.
 A: If you can factorize $x^2 - 8x + 7$, then you can solve a quadratic equation. 
Now recall that if the roots of $x^2+bx+c=0$ are $\alpha$ and $\beta$, then $-b = \alpha + \beta$ and $c = \alpha \beta$. 
(This is because $(x-\alpha)(x-\beta) = x^2 - (\alpha+\beta) x + \alpha \beta$.)
A: $x y = 7$
$x + y = -8$
$y = \frac{7}{x}$, so 
$x + \frac{7}{x} = -8$
Multiplying by $x$, yields:
$$x^2 +8x +7 = 0$$
$$(x+1)(x+7) = 0$$
Can you take it from there?
Regards
A: $(x-a)(x-b)=x^2-(a+b)x+ab$
So, you are looking for the roots of $x^2-(+2)x+(-3)$, that is, $x^2-2x-3$, which is $=x^2-2x+1-4=(x-1)^2-4$, and this is $0$ iff $(x-1)^2=4$, that is, $x-1=\pm 2$, i.e., $-1,3$ is your solution.
Similarly for the other one: $x^2-8x+7=(x-4)^2-16+7=\dots$
Or, the other way: $7+1=8$ and $7\cdot 1=7$, so the roots of $x^2-8x+7$ are $1$ and $7$.
A: I do not understand why you are trying to factorise $x^2-8x+7$. I suggest you use viete's formulae. xy=2 and x+y=-3. Let's say you have a quadratic equation $x^2+ax+b$ Then the roots $x_1, x_2$ has the property $x_1+x_2=-a$ and $x_1.x_2=c$ So you have the quadratic equation $x^2-2x-3$ When you solve this, you get the answer.
A: 
your questions answer are already users have given so i'm not repeating them.
  sometimes factorization is not easy because number are too big to manually calculate or may be fractional so try this formula as alternative.
  roots for equation $(ax^2+bx-c)=0$ is,
  $x=\dfrac{-b+\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ , 
  $\dfrac{-b-\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$ 

