How do you solve #x^2-5x+6=0 # by factoring?

Question

Eva Adamson · Accepted Answer

Since the #6# has a #+#-sign you're looking for two factors of #6# that add up to #-5# These can only be #-2and-3# so the factoring goes: #(x-2)(x-3)=0->x=2orx=3#

Jace Anderson · Answer

undefined To solve the quadratic equation $x^2 - 5x + 6 = 0$ by factoring:

1. Identify two numbers that multiply to the constant term $6$ and add up to the coefficient of the linear term $-5$.
2. Rewrite the middle term $ -5x $ using these two numbers.
3. Factor the quadratic expression into two binomial factors.
4. Set each binomial factor equal to zero and solve for $x$.

Following these steps:

1. The two numbers that multiply to $6$ and add up to $-5$ are $-2$ and $-3$.
2. Rewrite $ -5x $ as $-2x - 3x$.
3. Factor the quadratic expression: $x^2 - 2x - 3x + 6 = 0$ which factors as $(x - 2)(x - 3) = 0$.
4. Set each factor equal to zero:
   - $x - 2 = 0$ implies $x = 2$
   - $x - 3 = 0$ implies $x = 3$

So, the solutions to the equation $x^2 - 5x + 6 = 0$ are $x = 2$ and $x = 3$.

Avery Alexander · Answer

undefined To solve the equation x^2 - 5x + 6 = 0 by factoring, you need to find two numbers that multiply to give you 6 and add to give you -5 (which is the coefficient of the x term). These numbers are -2 and -3. Then, you rewrite the equation using these numbers to split the middle term:

x^2 - 2x - 3x + 6 = 0

Next, you group the terms:

(x^2 - 2x) + (-3x + 6) = 0

Now, you factor by grouping:

x(x - 2) - 3(x - 2) = 0

Now, you can see that (x - 2) is a common factor:

(x - 2)(x - 3) = 0

To find the solutions, you set each factor equal to zero:

x - 2 = 0
x - 3 = 0

Then solve for x:

x = 2
x = 3

Therefore, the solutions to the equation x^2 - 5x + 6 = 0 are x = 2 and x = 3.