Baka to Test to Shoukanjuu Season 1 Episode 1
2019/07/07
1. Factor the following expressions:
(1) a^2 b + ab^2 + a + b - ab - 1
(2) (x + y)(y + z)(z + x) + xyz
(3) x(y^2 - z^2) + y(z^2 - x^2) + z(x^2 - y^2)
(4) x^3 + y^3 + z^3 - 3xyz
(5) x^4 - 13x^2 + 36
(6) (x^2 + x - 5)(x^2 + x - 7) + 1
(7) (a + b + c + 1)(a + 1) + bc
(8) 4x^2 - 4xy + y^2 - 6x + 3y - 10
(9) a^5 + b^5 - a^3 b^2 - a^2 b^3
Answer:
TL;DR: Use Wolfram|Alpha.
= a^2 b + ab^2 - ab + a + b - 1 ← rearranged the terms
= ab(a + b - 1) + (a + b - 1) ← factored out ab from the first three terms= (ab + 1)(a + b - 1) ← factored out (a + b - 1)
= (x + y)(x + z)(y + z) + xyz ← rearranged the factors
= (x^2 + (y + z)x + yz)(y + z) + xyz ← expanded
= (y + z)x^2 + (y + z)^2 x + (y + z)yz + xyz ← expanded
= (x + (y + z))(y + z)x + ((y + z) + x)yz ← factored out (y + z)x from the first two terms, and factored out yz from the last two terms
= (x + y + z)(xy + xz + yz) ← factored out (x + y + z)
= x(y^2 - z^2) + yz^2 - yx^2 + zx^2 - zy^2 ← expanded the last two terms
= -(y - z)x^2 + (y^2 - z^2)x - y^2 z + yz^2 ← rearranged in descending powers of x
= -(y - z)x^2 + (y - z)(y + z)x - (y - z)yz ← from the difference of two squares= -(y - z)(x^2 - (y + z)x + yz) ← factored out -(y - z)
= -(y - z)(x - y)(x - z) ← from the quadratic factorization
= -(x - y)(x - z)(y - z) ← rearranged the factors
= (x + y)^3 - 3xy(x + y) + z^3 - 3xyz ← from the binomial theorem
= (x + y)^3 + z^3 - 3xy(x + y + z) ← rearranged the terms
= ((x + y) + z)^3 - 3(x + y)z((x + y) + z) - 3xy(x + y + z) ← from the binomial theorem
= (x + y + z)((x + y + z)^2 - 3(x + y)z - 3xy) ← factored out (x + y + z)= (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) ← expanded
(5) x^4 - 13x^2 + 36
= (x^2 - 4)(x^2 - 9) ← from the quadratic factorization
= (x + 2)(x - 2)(x + 3)(x - 3) ← from the difference of two squares
= (x^2 + x)^2 - 12(x^2 + x) + 36 ← from the quadratic factorization
= (x^2 + x - 6)^2 ← from the quadratic factorization= ((x + 3)(x - 2))^2 ← from the quadratic factorization
= (a + 1 + b + c)(a + 1) + bc ← rearranged the terms
= (a + 1)^2 + (b + c)(a + 1) + bc ← expanded
= (a + 1 + b)(a + 1 + c) ← from the quadratic factorization
= (a + b + 1)(a + c + 1) ← rearranged the terms
= (2x - y)^2 - 3(2x - y) - 10 ← from the binomial theorem
= (2x - y - 5)(2x - y + 2) ← from the quadratic factorization
= a^5 - a^3 b^2 - a^2 b^3 + b^5 ← rearranged the terms
= (a^2 - b^2)a^3 - (a^2 - b^2)b^3 ← factored out a^3 from the first two terms and factored out b^3 from the last two terms= (a^2 - b^2)(a^3 - b^3) ← factored out (a^2 - b^2)
= (a + b)(a - b)^2 (a^2 + ab + b^2) ← from the difference of two squares and two cubes
It is worth studying algorithms for factorization of polynomials.
(a + b)^n = sum_(k = 0)^n C(n, k) a^(n - k) b^k
a^n - b^n = (a - b) sum_(k = 0)^(n - 1) a^(n - k - 1) b^k
Sum of two nth powers (when n is odd):
a^n + b^n = (a + b) sum_(k = 0)^(n - 1) (-1)^k a^(n - k - 1) b^k
x^2 - (α + β)x + αβ = (x - α)(x - β)
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