Lesson 5 · from §52 of the 1910 text
Trinomials $x^2 + bx + c$
Suppose (x + h)(x + k) = x^2 + bx + c. Multiplying out, x^2 + (h + k)x + hk, so we need h + k = b \quad\text{and}\quad hk = c. If two such numbers exist the trinomial factors; if not, it's prime.
Rule. Find two numbers whose product is c and whose sum is b. They're the unlike terms of two binomials with common term x.
Examples. - x^2 + 7x + 12: pairs for 12 are 1 \cdot 12, 2 \cdot 6, 3 \cdot 4, and only 3 and 4 add to 7. So it's (x + 3)(x + 4). - x^2 - 12x + 32: the product is positive and the sum negative, so both numbers are negative: (x - 4)(x - 8). - x^2 - 6x - 40: a negative product means opposite signs, and the negative one is larger: (x + 4)(x - 10). - x^2 + 13x - 48 = (x - 3)(x + 16).
Quick check
x^2 - 2x - 24
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Factor.
x^2 + 5x + 6
x^2 + 9x + 20
x^2 - 7x + 12
x^2 - 9x + 14
x^2 + 19x + 18
x^2 - 18x + 17
x^2 - x - 6
x^2 - 3x - 40
x^2 + x - 30
x^2 - 2x - 35
x^2 + 11x - 42
6 - 5x - x^2
14 - 5x - x^2
21 + 4x - x^2
24 + 10a - a^2
10 + 3x - x^2
72 + c - c^2
x^2 + x - 90
28 + x^2 - 11x
27 - 12x + x^2
8x + x^2 - 48
x^2 - x - 90
x^2 - 5ax + 6a^2
x^2 + 7ax + 12a^2
a^2x^2 - 2ax - 15
h^2k^2 - 9hk + 20
x^4 - 5x^2 - 14
x^2 + x^4 - 110
x^6 - 11x^3 + 18
b^2x^2 - 3bxy - 28y^2
h^2k^2 - 5hkx - 36x^2
a^6b^2 - 5a^3bx - 24x^2
h^4d^2 + 10h^2dg^3 - 24g^6
9k^2x^4y^6 + k^4x^8 - 22y^12
Review: factor completely.
x^4 + 6x^3 + 9x^2
x^3 - 5x^2 + 6x
3b^2 - 9b - 30
3ax^3 - 3ax^2 - 6ax
4x^5 - 4x^3 - 360x
3x^4 - 21x^3 - 54x^2
4a^6y^4 - 64a^4y^2 + 256a^2
3x - 3ax - 18a^2x
56x + x^2 - x^3
140h^2 + 23h^3 - h^4
3x^3 - 3x^4 + 60x^2
32a^2x^2 + 2a^2 - 16a^2x
4ay^2x^2 - 40ay^2x + 100ay^2
3x^4 - 3x^3 + 3x^2 - 3x
2ax^3 - 4ax^2 - 12ax + 24a
5xyz - 15yz + 10xz - 30z