91Ó°ÊÓ

The given Polynomial is$49b^2-112b+64.$

Terms of a polynomial do not have any common factor

the polynomial is trinomial where first and last terms are perfect square

trinomial is in the pattern of $a^2-2ab+b^2$

Factorize it by using the formula for the square of the difference of two terms.

$$\begin{gathered} 49b^2-112b+64={\left(7b\right)}^2-112b+8^2 \\ ={\left(7b\right)}^2-2·7b·8+8^2 \\ ={(7b-8)}^2 \end{gathered}$$

Polynomial cannot be factored further.

So factored form islocalid="1647945678602" $49b^2-112b+64={(7b-8)}^2.$

Check whether the expanded form of the factored polynomial is equal to the original polynomial or not.

$$\begin{gathered} 49b^2-112b+64\stackrel{?}{=}{(7b-8)}^2 \\ 49b^2-112b+64\stackrel{?}{=}(7b-8)(7b-8) \\ 49b^2-112b+64\stackrel{?}{=}49b^2-56b-56b+64 \\ 49b^2-112b+64\stackrel{?}{=}{\left(7b\right)}^2-112b+8^2 \\ 49b^2-112b+64=49b^2-112b+64 \end{gathered}$$

The left-hand side is equal to the right-hand side so factored form is correct.