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Question: Factorize \[y^{3} + y^{2} - y - 1\] A.\[(y + 1)(y + 2)(y - 1)\] B.\[\left( y + 1 \right)^{2}(y -...

Factorize y3+y2−y−1y^{3} + y^{2} - y - 1
A.(y+1)(y+2)(y−1)(y + 1)(y + 2)(y - 1)
B.(y+1)2(y−1) \left( y + 1 \right)^{2}(y - 1)\
C. (y−1)2(y+1) \ \left( y - 1 \right)^{2}(y + 1)\
D.None of these{None\ of\ these}

Explanation

Solution

We need to factorize the given expression. Factorization is nothing but writing a whole number into smaller numbers of the same kind. By using algebraic formulas and also by taking the common terms outside, we can factorize the given expression.
Formula used :
a2–b2=(a+b)(a–b)a^{2} – b^{2} = (a + b)(a – b)

Complete answer:
Let us consider the given expression as
f(y)=y3+y2–y–1f\left( y \right) = y^{3} + y^{2} – y – 1
By taking y2y^{2} common from the first two terms,
f(y)=y2(y+1)–y–1f\left( y \right) = y^{2}\left( y + 1 \right) – y – 1
Then now by taking (−1)( -1) from last two terms,
We get
f(y)=y2(y+1)–1(y+1)f\left( y \right) = y^{2}\left( y + 1 \right) – 1\left( y + 1 \right)
Now taking (y+1)(y+1) common,
We get,
f(y)=(y2–1)(y+1)f\left( y \right) = {(y}^{2} – 1)(y + 1)
Now here
(y2–1)\left( y^{2} – 1 \right) is in the form of (a2–b2)\left( a^{2} – b^{2} \right)
With the help of the formula we can expand the expression,
(y2–1)=(y+1)(y–1)\left( y^{2} – 1 \right) = \left( y + 1 \right)\left( y – 1 \right)
Now we get,
f(y)=(y+1)(y–1)(y+1)f\left( y \right) = \left( y + 1 \right)\left( y – 1 \right)\left( y + 1 \right)
By multiplying,
We get,
f(y)=(y+1)2(y–1)f\left( y \right) = \left( y + 1 \right)^{2}\left( y – 1 \right)
Final answer :
Thus the answer is (y+1)2(y–1)\left( y + 1 \right)^{2}\left( y – 1 \right)
Option. B) (y+1)2(y–1)\left( y + 1 \right)^{2}\left( y – 1 \right)

Additional information
The method to find the factor of the integers is an easy method but we can’t factorize the algebraic equation in that manner.
Basic formula used in factorization are
a2−b2= (a − b)(a + b)a^{2}-b^{2} = \ (a\ - \ b)(a\ + \ b)
(a + b)2= a2+ 2ab + b2\left( a\ + \ b \right)^{2} = \ a^{2} + \ 2ab\ + \ b^{2}
(a + b)2 = a2− 2ab + b2\left( a\ + \ b \right)^{2}\ = \ a^{2} - \ 2ab\ + \ b^{2}

Note:
In other words, factorization is known as the decomposition of the mathematical objects to the product of smaller objects. Matrices also possess the process of factorization. The formula of factorization is N=Xa×Yb×ZcN = X^{a} \times Y^{b} \times Z^{c}
Where a,b,ca, b, c are the exponential powers of the factor. There are five methods in factorization. We can reduce any algebraic expressions into smaller objects where the equations are represented as the product of factors.