Factor each of following expressions
x (4th power) - x (3rd power) - 6x (2nd power
(a-b)(2nd power) - (a 2nd power + b) 2nd power
x^2(x^2-x-6)
= x^2 (x-3)(x+2)
(a-b)^2 - (a^2+b)^2
a^2-2ab+b^2 -(a^4 + 2 a^2 b + b^2)
-a^4 - 2 a^2 b - 2 a b
-a (a^3 + 2 a b + 2 b)
I can not see more to do
To factor each expression, we need to identify common factors and apply factoring techniques. Let's factor each expression one by one:
1. x^4 - x^3 - 6x^2
First, let's look for common factors among the terms. In this case, we can factor out an 'x^2' since it is a common factor in all three terms:
x^4 - x^3 - 6x^2 = x^2(x^2 - x - 6)
Now, we can focus on factoring the quadratic expression inside the parentheses, x^2 - x - 6. To factor it further, we need to find two numbers whose product is -6 and sum is -1 (since -1 is the coefficient of 'x').
The numbers that satisfy this condition are -3 and 2. So, we rewrite the quadratic expression as:
x^2 - x - 6 = (x - 3)(x + 2)
Therefore, the fully factored expression is:
x^4 - x^3 - 6x^2 = x^2(x - 3)(x + 2)
2. (a - b)^2 - (a^2 + b)^2
In this expression, we have the difference of squares and can use the algebraic identity:
(a - b)^2 - (a^2 + b)^2 = (a - b + a^2 + b)(a - b - a^2 - b)
Simplifying further, we get:
(a - b + a^2 + b)(a - b - a^2 - b) = (a^2 + a - b + b)(a - b - a^2 - b)
Since (+b - b) cancels out and (-a^2 - a + a^2) cancels out, the expression becomes:
(a^2 + a)(a - b - a^2 - b)
Therefore, the fully factored expression is:
(a - b)^2 - (a^2 + b)^2 = (a^2 + a)(a - b - a^2 - b)