Question

One of the factors of $$\left(16y^2-1\right)+{\left(1-4y\right)}^2$$ is
A.$$\left(4+y\right)$$
B.$$\left(4-y\right)$$
C.$$\left(4y+1\right)$$
D.$$8y$$

Answer 1

Hint: Here we will first use the algebraic identity to expand the square terms given in the expression. Then we will simplify and solve the equation formed. We will then take the common terms from the final equation to get the factors of the given expression.

Complete step-by-step answer:
Given expression is $$\left(16y^2-1\right)+{\left(1-4y\right)}^2$$.
First, we will expand the squares of the given expression by using the basic algebraic identity i.e. $${\left(a-b\right)}^2=a^2-2ab+b^2$$ in the given expression. Therefore, we get
$$\left(16y^2-1\right)+{\left(1-4y\right)}^2=\left(16y^2-1\right)+\left(1^2+{\left(4y\right)}^2-\left(2\times 1\times 4y\right)\right)$$
Applying the exponent and multiplying the terms, we get
$$\Rightarrow \left(16y^2-1\right)+{\left(1-4y\right)}^2=\left(16y^2-1\right)+\left(1+16y^2-8y\right)$$
Adding and subtracting the like terms, we get
$$\Rightarrow \left(16y^2-1\right)+{\left(1-4y\right)}^2=32y^2-8y$$
Now we will take $$8y$$ common from both the terms in the above equation to form the factors of the equation. Therefore, we get
 $$\Rightarrow \left(16y^2-1\right)+{\left(1-4y\right)}^2=8y\left(4y-1\right)$$
Hence, $$8y$$ and $$\left(4y-1\right)$$ are the two factors of the given expression i.e. $$\left(16y^2-1\right)+{\left(1-4y\right)}^2$$.
So, option D is the correct option.

Note: Factors are the smallest numbers with which the given number is divisible and their multiplication will give the original number. The process of the formation of the factors is generally known as factorization. Factorization is the process in which a number is written in the forms of its small factors which on multiplication give the original number. Here we have to note that for expanding the square equation we should know the basic algebraic identities. Algebraic identities are equations where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation.