How to factorize 4a to the power 4 + 81 b to the power 4 ?
Please help me
Answers
A4 – b4 = (a – b)(a3 + a2b + ab2 + b3) = (a – b) [ a2(a + b) + b2(a + b) ] = (a – b)(a + b)(a2 + b2) or, more simply, a4 – b4 = (a2 – b2)(a2 + b2) = (a – b)(a + b)(a2 + b2)
questions b.
p^4-81=0
p^4=81
p=(81)^1/4
p=3
Every polynomial that is a difference of squares can be factored by applying the following formula:
\blueD{a}^2-\greenD{b}^2=(\blueD a+\greenD b)(\blueD a-\greenD b)a
2
−b
2
=(a+b)(a−b)start color #11accd, a, end color #11accd, squared, minus, start color #1fab54, b, end color #1fab54, squared, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis
Note that aaa and bbb in the pattern can be any algebraic expression. For example, for a=xa=xa, equals, x and b=2b=2b, equals, 2, we get the following:
\begin{aligned}\blueD{x}^2-\greenD{2}^2=(\blueD x+\greenD 2)(\blueD x-\greenD 2)\end{aligned}
x
2
−2
2
=(x+2)(x−2)
The polynomial x^2-4x
2
−4x, squared, minus, 4 is now expressed in factored form, (x+2)(x-2)(x+2)(x−2)left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 2, right parenthesis. We can expand the right-hand side of this equation to justify the factorization:
\begin{aligned}(x+2)(x-2)&=x(x-2)+2(x-2)\\\\&=x^2-2x+2x-4\\ \\ &=x^2-4\end{aligned}
(x+2)(x−2)
=x(x−2)+2(x−2)
=x
2
−2x+2x−4
=x
2
−4
Now that we understand the pattern, let's use it to factor a few more polynomials.
Example 1: Factoring x^2-16x
2
−16x, squared, minus, 16
Both x^2x
2
x, squared and 161616 are perfect squares, since x^2=(\blueD{x})^2x
2
=(x)
2
x, squared, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared and 16=(\greenD{4})^216=(4)
2
16, equals, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared. In other words:
x^2-16 =(\blueD {x})^2-(\greenD{4})^2x
2
−16=(x)
2
−(4)
2
x, squared, minus, 16, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared, minus, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared
Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression:
\blueD{a}^2-\greenD{b}^2=(\blueD a+\greenD b)(\blueD a-\greenD b)a
2
−b
2
=(a+b)(a−b)start color #11accd, a, end color #11accd, squared, minus, start color #1fab54, b, end color #1fab54, squared, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis
In our case, \blueD a=\blueD xa=xstart color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd and \greenD b=\greenD 4b=4start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 4, end color #1fab54. Therefore, our polynomial factors as follows:
(\blueD{x})^2-(\greenD{4})^2=(\blueD x+\greenD 4)(\blueD x-\greenD 4)(x)
2
−(4)
2