Question 4: The HCF of two polynomials is \(y-7\) and their LCM is \(y^3-10y^2+11y+70\). If one of the polynomials is \(y^2-5y-14\), find the other.
Solution:
Since we know that $\displaystyle p( x) \times q( x) =HCF\times LCM$
Let $\displaystyle p( x)$ is our required polynomial, we have
$\displaystyle p( x) \times y^{2} -5y-14=y-7\times$$y^{3} -10y^{2} +11y+70$
$\displaystyle p( x) =\frac{( y-7)\left( y^{3} -10y^{2} +11y+70\right)}{y^{2} -5y-14}$
$\displaystyle \ \ \ \ \ \ \ =\frac{( y-7)\left( y^{3} -10y^{2} +11y+70\right)}{y^{2} -7y+2y-14}$
$\displaystyle \ \ \ \ \ \ \ =\frac{( y-7)\left( y^{3} -10y^{2} +11y+70\right)}{y( y -7) +2( y-7)}$
$\displaystyle \ \ \ \ \ \ \ =\frac{( y-7)\left( y^{3} -10y^{2} +11y+70\right)}{( y -7)( y+2)}$
$\displaystyle \ \ \ \ \ \ \ =\frac{y^{3} -10y^{2} +11y+70}{y+2}$
$\begin{array}{ r l }
& \ \underline{\ \ \ \ \ \ y^{2} -12y+35\ \ \ \ \ \ \ \ }\\
y+2 & \bigr) \ y^{3} -10y^{2} +11y+70\\
& \underline{\underset{\ \ -}{} y^{3} \ \underset{-}{+} \ 2y^{2} \ \ \ \ \ \ \ \ \ }\\
& \ \ \ \ \ \ \ \ -12y^{2} +11y+70\\
& \ \ \ \ \ \ \ \ \underline{\underset{+}{-} 12y^{2} \ \underset{+}{-} \ 24y\ \ \ \ } \ \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35y+70\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\underset{-}{} 35y\ \underset{-}{+} 70\ }\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
Our required polynomial is $y^{2} -12y+35$