When evaluating a quadratic equation, we want to find two x values that make the equation equal to 0. To factor, seek for two integers that multiply to produce +6 in the constant term (the phrase without any x). These are the numbers +1 and +6, -1 and -6, +2 and +3, and -2 and -3. Now we go at the x term in the equation, +5x, to see which of these pairings adds up to +5. This equals +2 and +3. As a result, the factorisation of the equation is (x +2)(x +3) = 0, which can be verified by multiplying out the brackets to get the original equation (for example, using the FOIL method.) The problem may now be solved by determining the two x values that make each of the two brackets equal to 0. This is equivalent to solving x+2=0 and x+3=0 equations. As a result, the solutions to this equation are x=-2 and x=-3.

### Solve By Factorization X2+5X+6=0

This equation is written in standard form: ax2+bx+c=0. . In the quadratic formula, rac-bsqrtb2-4ac2a, substitute 1 for a, 5 for b, and 6 for c. All equations using the formula ax2+bx+c=0 may be solved using the quadratic formula: rac-bsqrtb2-4ac2a The quadratic formula provides two answers, one for addition and one for subtraction.

### Solve By Factorization Method X2-5X+6=0

This sort of issue is a freebie in my opinion since it is already set up for us to solve. Take note that the left side includes polynomial factors, whereas the right side is just zero! All we have to do is set each element to zero and solve each equation for x.