kyotsu-test 2016 QCourse1-II-Q1

kyotsu-test · Japan · eju-math__session1 Polynomial Division & Manipulation
We are to find the value of $a$ such that $15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a$ can be factorized as the product of linear expressions in $x$ and $y$.
First of all, the first three terms of the above expression form a quadratic expression in $x$ and $y$ that can be factorized as
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } = ( \mathbf { A } x - \mathbf { B } y ) ( \mathbf { C } x + \mathbf { D } y ) .$$
Hence, when we set
$$\begin{aligned} & 15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a \\ & \quad = ( \mathbf { A } x - \mathbf { B } y + b ) ( \mathbf { C } x + \mathbf { D } y + c ) , \end{aligned}$$
the right-hand side of equation (1) can be expanded into
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } + ( \mathbf { E } b + \mathbf { F } c ) x + ( \mathbf { G } b - \mathbf { H } c ) y + b c .$$
When we compare the coefficients of this expression with the coefficients of the left-hand side of equation (1), we have
$$b = \mathbf { I } , \quad c = - \mathbf { J } ,$$
and hence $a = - \mathbf { K L }$.
We are to find the value of $a$ such that $15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a$ can be factorized as the product of linear expressions in $x$ and $y$.

First of all, the first three terms of the above expression form a quadratic expression in $x$ and $y$ that can be factorized as

$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } = ( \mathbf { A } x - \mathbf { B } y ) ( \mathbf { C } x + \mathbf { D } y ) .$$

Hence, when we set

$$\begin{aligned}
& 15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a \\
& \quad = ( \mathbf { A } x - \mathbf { B } y + b ) ( \mathbf { C } x + \mathbf { D } y + c ) ,
\end{aligned}$$

the right-hand side of equation (1) can be expanded into

$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } + ( \mathbf { E } b + \mathbf { F } c ) x + ( \mathbf { G } b - \mathbf { H } c ) y + b c .$$

When we compare the coefficients of this expression with the coefficients of the left-hand side of equation (1), we have

$$b = \mathbf { I } , \quad c = - \mathbf { J } ,$$

and hence $a = - \mathbf { K L }$.