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I need to find $2$ factors of $-357$, which add up to $4$. Obviously one number is positive and the other is negative. I understand this and I know the factors can be $21$ and $-17$; but, how do I factor larger numbers like $-357$ without using a guess/check method?

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  • $\begingroup$ No, the roots are not $21$ and $-17$ and they have to add up to $-4$, not $4$. The answer is actually $-21$ and $17$. $\endgroup$
    – user26486
    Mar 22, 2015 at 18:44
  • $\begingroup$ Why? 4 is the coefficient of 4x $\endgroup$
    – Ben
    Mar 22, 2015 at 18:45
  • $\begingroup$ Vieta's formulas say that the roots of $x^2+bx+c$ add up to $-b$ (and the product of the roots is $c$). $\endgroup$
    – user26486
    Mar 22, 2015 at 18:47

3 Answers 3

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You could just solve it using the quadratic formula. Then, if $x_1$ and $x_2$ are the two solutions, you can factor this as $x^2+4x-357=(x-x_1)(x-x_2)$.


If you really want to factorize $357$, though, you could just test the small primes to see if they are factors. For instance, $2$ is clearly NOT a factor, but $3$ is (because if you add up all of the digits you get a number which $3$ divides). So you factor out a $3$ to get $357=3\cdot(119)$.

Then checking $119$, $2$, $3$, and $5$ aren't factors -- but $7$ is. So then you have $357=3\cdot7\cdot17$. And $17$ is prime. So that's how you factor $357$.

Rules for Testing Small Primes: Given an integer we want to factor, here are some rules for checking whether the number is divisible by the small primes:

  1. The number will be divisible by $2$ if the last digit of the number is even.
  2. The number will be divisible by $3$ if the sum of the digits is divisible by $3$. If you can't immediately tell if the sum of the digits is divisible by $3$, add the digits again (and repeat as many times as necessary).
  3. The number will be divisible by $5$ if its last digit is a $0$ or a $5$.
  4. The number will be divisible by $7$ if after you double the last digits and subtract that number from the truncated number, you end up with a number divisible by $7$. For example: we test $1617$ by subtracting $2\cdot 7$ from $161$ to get $147$. If you can't immediately tell that this is divisible by $7$, then let's do it again: $14-2\cdot 7 =0$. $0$ is divisible by $7$ (I know it's weird, but $0$ is actually divisible by ALL nonzero integers), so $1617$ IS divisible by $7$.
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By completing the square

$$x^2 + 4x - 357 = 0 \implies (x + 2)^2 - 4 - 357 = 0 \implies (x+2)^2 = 361 \implies x= -2 \pm 19$$

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The formula to get solutions for $ax^2+bx+c$ is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, otherwise you must trial and error only with each of its factors (of 357).

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