5.3 Q-4

Exercise 5.3 — Question 4

Problem Statement

Factorize the polynomial $P(y)=y^3-2y^2-y+2$ completely using the Factor Theorem.

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Key Concepts

Remainder Theorem: When a polynomial $P(x)$ is divided by $(x-k)$, the remainder equals $P(k)$.

Factor Theorem: $(x-k)$ is a factor of $P(x)$ if and only if $P(k)=0$.

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Step-by-Step Solution

Step 1: Find a rational root by testing factors of the constant term.

The constant term is $2$. Possible rational roots: $\pm 1,\pm 2$.

Test $y=1$: $P(1)=(1{)}^3-2(1{)}^2-(1)+2=1-2-1+2=0$

Since $P(1)=0$, by the Factor Theorem, $(y-1)$ is a factor.

Step 2: Divide $P(y)$ by $(y-1)$ using synthetic division.

$\begin{array}{rrrrr}1& 1& -2& -1& 2\\ & & 1& -1& -2\\ & 1& -1& -2& 0\end{array}$

Quotient: $y^2-y-2$, Remainder: $0$

So: $P(y)=(y-1)(y^2-y-2)$

Step 3: Factorize the quadratic $y^2-y-2$.

$y^2-y-2=(y-2)(y+1)$

Verification: $(y-2)(y+1)=y^2+y-2y-2=y^2-y-2$ ✓

Step 4: Write the complete factorization.

$\boxed{P(y)=(y-1)(y-2)(y+1)}$

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Verification Using Factor Theorem

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Real-World Application

If a rectangular solid has volume $V(y)=y^3-2y^2-y+2$ and one dimension is $(y-2)$, then:

$V(y)=(y-2)(y-1)(y+1)$

The three dimensions are $(y-2)$, $(y-1)$, and $(y+1)$.

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