Can you factorise it?

Given that $(a_1x + a_2y + a_3)(a_4x + a_5y + a_6) = 3x^2-16xy + 16y^2 + 17x - 44y + 24$ , equating the coefficients we have:

$\begin{cases} \begin{aligned} a_1 a_4 & = 3 & ...(1) \\ a_1a_5 + a_2a_4 & = -16 & ...(2) \\ a_2a_5 & = 16 & ...(3) \\ a_1a_6 + a_3a_4 & = 17 & ...(4) \\ a_2a_6 + a_3a_5 & = -44 & ...(5) \\ a_3a_6 & = 24 &...(6) \end{aligned} \end{cases}$

With six equations, we can solve the six unknown $a_k$ , but it is very tedious. We can just make some educated guesses.

• Since $(1): \ a_1a_4 = 3$ , we can assume $a_1 = 3$ and $a_4 = 1$ .
• Next $(3): \ y^2$ , $a_2 a_5 = 16$ . We note that when both $a_2$ and $a_5$ appear in $(2)$ and $(5)$ , the coefficients are negative. Therefore, we can assume $a_2 = a_5 = -4$ .
• Then $(2):\ (3)(-4)+(1)(-4) = -16$ , which is correct.
• And $(4)$ and $(5)$ become:

$\quad \begin{cases} \begin{aligned} (4): \quad \ \ 3a_6 + a_3 & = 17 & ...(4a) \\ (5): \ -4a_6 - 4a_3 & = -44 \\ a_6 + a_3 & = 11 &...(5a) \end{aligned} \end{cases} \implies (4a) - (5a): \ 2a_6 = 6 \implies a_6 = 3 \implies a_3 = 8$

Therefore $a_1+a_2+a_3+a_4+a_5+a_6 = 3-4+8+1-4+3 = \boxed 7$ .

$\begin{aligned} \text{Expression} &= \underbrace{3x^2 -16xy + 16y^2}_{\text{Degree 2}} + \underbrace{17x -44y +24}_{\text{Degree 1}} \\ &= \bigg( 3x^2 -12xy -4xy + 16y^2 \bigg) + \bigg( 17x -44y +24 \bigg) \\ &= \bigg( 3x(x-4y) -4y(x-4y) \bigg) + \bigg( 17x -44y +24 \bigg) \\ &= (3x-4y)(x-4y) + \bigg( 17x -44y +24 \bigg) \\ (3x-4y + a_{3})(x-4y + a_{6}) &= (3x-4y)(x-4y) + \bigg( 17x -44y +24 \bigg) \\ \\ & \text{Comparing coefficients} \\ \\ a_{3}(x) + a_{6}(3x) = 17x &\implies a_{3} + 3a_{6} = 17 \ldots (1) \\ a_{3}(-4y) + a_{6}(-4y) = -44y &\implies a_{3} + a_{6} = 11 \ldots (2) \\ \\ \text{Solving coefficients} &\implies a_{3} = 8, a_{6} = 3 \\ \\ 3x^2 -16xy + 16y^2 +17x -44y +24 &= \boxed{(3x-4y + 8)(x-4y + 3)} \end{aligned}$

$3-4+8+1-4+3 = \color{#D61F06}{\boxed{7}}$

How do I post an image 😅 I solved the problem but don't know how to post image😅.

You can use the third button when on editing mode.