Question

H.C.F of $x^2+5x+6$ and $x^3+27$ is:

• A. $x+2$
• B. $x-2$
• C. $x-3$
• D. $x+3$

Answer 1

The correct option is D

$x+3$

Solution:

Step 1: Finding the factors of $\mathit{p}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{:}$

Given: Let , $p\left(x\right)=x^2+5x+6$ and $q\left(x\right)=x^3+27$.Given: Let , $p\left(x\right)=x^2+5x+6$ and $q\left(x\right)=x^3+27$.

$p\left(x\right)=x^2+5x+6$

$$\begin{gathered} =x^2+2x+3x+6\left[\mathrm{splitting}\mathrm{the}\mathrm{middle}\mathrm{term}\right] \\ =x(x+2)+3(x+2) \\ =(x+2)(x+3) \end{gathered}$$

Step 2: Finding the factors of $\mathit{q}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{:}$

$$\begin{gathered} q\left(x\right)=x^3+27 \\ =x^3+3^3 \\ =(x+3)(x^2-3x+9) \end{gathered}$$ $\left[\because a^3+b^3=\left(a+b\right)\left(a^2-\mathrm{ab}+b^2\right)\right]$

Step 3: Finding the HCF:

We found that the factors of $p\left(x\right)$ are $(x+2)\&(x+3)$ and the factors of $q\left(x\right)$ are $(x+3)\&(x^2-3x+9)$. So the highest common factor is $(x+3)$.

Therefore, H.C.F is $\mathbf{(}\mathit{x}\mathbf{+}\mathbf{3}\mathbf{)}$.

Final answer: Hence, option (D) is the correct one.