Question

factorice and divide -b^3(1-256b^4) ÷ b^2(1-4b)​

Answer 1

The factorization is $-b(1+16b^{2})(1+4b^{})$.

• The product of a polynomial's factors with degrees that are less than or equal to the original polynomial can be used to represent a polynomial. Factorization of polynomials is the technical term for the factoring process. The process of factoring a number into positive prime factors is a common one. A prime number is one with only 1 and itself as its positive components. There is just one way to factorise an integer into positive prime components.
• To obtain the specified polynomial, all the terms that were multiplied must be known. Then, until you are unable to factor any more, try to factor each term you obtained in the first step. The polynomial is considered to be factored entirely when there is no more factoring left to do.

Here, the expression is given as,

$\frac{-b^{3}(1-256b^{4}) }{b^{2}(1-4b) }$

$\frac{-b(1+16b^{2})(1-16b^{2}) }{(1-4b) }\\=\frac{-b(1+16b^{2})(1+4b^{}) (1-4b) }{(1-4b) }\\\\=-b(1+16b^{2})(1+4b^{})$

Hence, the factorization is $-b(1+16b^{2})(1+4b^{})$.

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