Lesson 16 · from §96 of the 1910 text
Factors Involving Radicals
Factors involving radicals
In the chapter on factoring, factors with radicals were not allowed. That limit is no longer necessary, so many expressions that were prime before can now be factored: 3x^2 - 1 = \left(\sqrt3\,x + 1\right)\left(\sqrt3\,x - 1\right), \qquad 4x^2 - 5 = \left(2x + \sqrt5\right)\left(2x - \sqrt5\right).
The variable is not allowed under a radical sign in a factor. So although \left(x + \sqrt x + 1\right)\left(x - \sqrt x + 1\right) = x^2 + x + 1, the trinomial x^2 + x + 1 is not regarded as factorable this way. In this extension of our notion of a factor, only the coefficients may contain radicals.
The sum and difference of two cubes work the same way: x^3 + 5 = x^3 + \left(\sqrt[3]5\right)^3 = \left(x + \sqrt[3]5\right)\left(x^2 - \sqrt[3]5\,x + \sqrt[3]{25}\right).
Example. Solve x^2 - 7 = 0 by factoring.
\left(x + \sqrt7\right)\left(x - \sqrt7\right) = 0, so x + \sqrt7 = 0, giving x = -\sqrt7, or x - \sqrt7 = 0, giving x = \sqrt7. These values check at once in x^2 - 7 = 0: if a root is put in for the variable, one of the factors becomes zero.
Typing answers: factors like
(sqrt(3)x + 1)(sqrt(3)x - 1)or(x + cbrt(5))(x^2 - cbrt(5)x + cbrt(25)); factor completely, keep each radical in simplest form, and keep x out from under the radical signs. Roots are separated by commas, and \pm may be used:±sqrt(2), ±3.
From the history of algebra
François Vieta (1540–1603). Algebra is a universal language because its symbols for operations and relations are widely understood and adopted. This was not always so: for a long time there was no accepted notation, and each writer used any symbols that suited him. One of the men who did most to establish a fixed notation was François Vieta, a French lawyer who studied and wrote on mathematics as a pastime. He was in public life all his career and was well known for his ability to decipher captured enemy dispatches.
Vieta established the signs + and - for addition and subtraction, which had been used before him but were not generally accepted. He denoted the known numbers in an equation by the consonants B, C, D, \ldots and the unknowns by the vowels A, E, I, \ldots He recognized negative roots of equations but rejected them as absurd. For the second and third powers of the unknown he used Q (quadratus) and C (cubus), and instead of = he wrote aeq. (aequalis). So he would have written x^3 - 8x^2 + 16x = 40 as
1C - 8Q + 16N \text{ aeq. } 40,
while writers only a little earlier would have written Cubus m 8 Census p 16 rebus aequatur 40, a form in which operations are very hard to carry out.
Vieta was also the first to obtain an exact numerical expression for \pi. It calls for an infinite number of operations which could never all be performed, though the further one goes, the closer the approximation. In a sense the familiar sign \sqrt{\phantom{x}} also implies an infinite number of operations, since the square root of 2 never comes out even; but Vieta's expression for \pi was more involved and used complicated irrational fractions.
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} - 6
3 x^{2} - 4
5 x^{2} - 1
x^{4} - 4
4 x^{4} - 1
x^{3} - 2
x^{3} + 6
3 x^{3} - 1
x^{3} + 4
2 x^{3} + 8
2 x^{3} - 8
6 x^{3} + 24
Solve by factoring and check:
x^{2} - 2 = 0
x^{2} - 6 = 0
2 x^{2} - 1 = 0
x^{4} + 6 = 5 x^{2}
4 x^{4} + 5 = 12 x^{2}
3 x^{4} + 8 = 14 x^{2}
5 x^{4} - 16 x^{2} + 3 = 0
x^{4} + 8 a = 4 x^{2} + 2 a x^{2}
a x^{4} - x^{2} + 3 a = 3 a^{2} x^{2}
4 x^{4} + a = x^{2} + 4 a x^{2}