Expanding a quadratic equation (x-3)(x-2)=0

Focus on the left hand side of the equation: (x-3)(x-2) These are two factors, basically that (x-3) is a number, and (x-2) is another number. (x-2)(x-3) is just showing that you're multiplying these numbers together. To multiply these numbers and hence expand the quadratic equation, multiply each bit of it individually. You can do it in a specific order to remember it easily: First: the first numbers of both of the brackets. Outer: the outer numbers of the brackets. Inner: the inner numbers of the brackets. Last: the last numbers which haven't been multiplied yet. This makes the acronym FOIL. In this case, the First is x * x = x^2. So start with x^2. Now add the Outer numbers, which are -3 * -2 = 6. In total now we have x^2 + 6. The Inner numbers are -3 * x = -3x. Add this to the total to give x^2 - 3x + 6. Finally, the Last numbers are x * -2 = -2x. Giving the total as: x^2 - 3x - 2x + 6 = x^2 - 5x + 6

Remember that this is only the left hand side of the equation, and that it equals 0, this doesn't change because you have expanded it. The expanded form of this equation means the exact same as the factorised form. Now the answer is: x^2 - 5x + 6 = 0

DW
Answered by Daniel W. Maths tutor

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