Problema Solution
Explain how to factor the following trinomials forms: x2 + bx + c and ax2 + bx + c. Is there more than one way to factor this? Show your answer using both words and mathematical notation. Then post two problems for the class to factor.
Answer provided by our tutors
Factoring x^2 + bx + c:
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To factor x^2 + bx + c, find integers m, and n such that
x^2 + bx + c = (x-m)(x-n)
then we can write it as
x^2 + bx + c = x^2 - (m+n)x + mn
So the sum of m and n should be equal to - b and product of m and n should be equal to c.
m + n = -b
mn = c
m and n are the roots of the quadratic equation x^2 + bx + c = 0
Factoring ax^2 + bx + c:
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To factor ax^2 + bx + c, a ≠ 1,
ax^2 + bx + c = a(x^2 + (b/a)x + c/a)
find integers m and n such that
x^2 + (b/a)x + c/a = (x-m)(x-n) = x^2 - (m+n)x + mn
m+n = -b/a
mn = c/a
and finally we have
ax^2 + bx + c = a(x^2 + (b/a)x + c/a) = a(x-m)(x-n)
ax^2 + bx + c = a(x-m)(x-n)
m and n are the roots of the quadratic equation x^2 + (b/a)x + c/a = 0
Example1:
x^2 + 6x + 6
the roots of x^2 + 5x + 6 = 0 are m = -2, n=-3
m + n = -2 - 3 = -5
mn = (-2)(-3) = 6
click here to see the step by step solution
x^2 + 6x + 6 = (x-(-2))(x-(-3)) = (x+2)(x+3)
Example2:
3x^2 - 3x - 6
3x^2 - 3x - 6 = 3 (x^2 - x - 2)
the roots of x^2 - x - 2 = 0 are m= -1, n=2
m + n = 1 = -(-1)
mn = 02
click here to see the step by step solution
x^2 - x - 2 = (x-(-1))(x-2)=(x+1)(x-2)
3x^2 - 3x - 6 = 3 (x^2 - x - 2) = 3(x+1)(x-2)