1. **Factorize each expression completely:** (1) Factorize $x^2 - 1$. - This is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$. - Here, $a = x$, $b = 1$. - So, $x^2 - 1 = (x-1)(x+1)$. (2) Factorize $16x^2 - 9$. - This is also a difference of squares: $16x^2 = (4x)^2$, $9 = 3^2$. - So, $16x^2 - 9 = (4x - 3)(4x + 3)$. (3) Factorize $x^3 + 1$. - This is a sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. - Here, $a = x$, $b = 1$. - So, $x^3 + 1 = (x + 1)(x^2 - x + 1)$. (4) Factorize $27 - a^3 b^3$. - Rewrite as $27 - (ab)^3$. - This is a difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. - Here, $a = 3$, $b = ab$. - So, $27 - a^3 b^3 = (3 - ab)(9 + 3ab + a^2 b^2)$. (5) Factorize $4x^4 + 4000x$. - First, factor out the greatest common factor (GCF): $4x$. - $4x^4 + 4000x = 4x(x^3 + 1000)$. - Recognize $x^3 + 1000$ as sum of cubes: $x^3 + 10^3$. - So, $x^3 + 1000 = (x + 10)(x^2 - 10x + 100)$. - Final factorization: $4x(x + 10)(x^2 - 10x + 100)$. 2. **Given equations:** $$4x^2 - 9y^2 = 115$$ $$2x - 3y = 5$$ Find $2x + 3y$. - From the second equation, square both sides: $$ (2x - 3y)^2 = 5^2 = 25 $$ - Expand left side: $$ 4x^2 - 12xy + 9y^2 = 25 $$ - From the first equation: $$ 4x^2 - 9y^2 = 115 $$ - Add the two equations: $$ (4x^2 - 9y^2) + (4x^2 - 12xy + 9y^2) = 115 + 25 $$ $$ 8x^2 - 12xy = 140 $$ - Simplify: $$ 4x^2 - 6xy = 70 $$ - Now consider $(2x + 3y)^2$: $$ (2x + 3y)^2 = 4x^2 + 12xy + 9y^2 $$ - Using the first equation, rewrite $4x^2 + 9y^2$: $$ 4x^2 + 9y^2 = 115 + 2 imes 9y^2 = 115 + 18y^2 $$ - But this is complicated; instead, use the identity: $$ (2x + 3y)^2 + (2x - 3y)^2 = 2(4x^2 + 9y^2) $$ - Substitute known values: $$ (2x + 3y)^2 + 25 = 2 imes 115 = 230 $$ - So: $$ (2x + 3y)^2 = 230 - 25 = 205 $$ - Therefore: $$ 2x + 3y = \pm \sqrt{205} $$ 3. **Solve the equation:** $$ x^2 + 6x = -9 $$ - Move all terms to one side: $$ x^2 + 6x + 9 = 0 $$ - Recognize perfect square: $$ (x + 3)^2 = 0 $$ - So, solution is: $$ x = -3 $$ **Final answers:** (1) $(x-1)(x+1)$ (2) $(4x-3)(4x+3)$ (3) $(x+1)(x^2 - x + 1)$ (4) $(3 - ab)(9 + 3ab + a^2 b^2)$ (5) $4x(x + 10)(x^2 - 10x + 100)$ For Q4: $2x + 3y = \pm \sqrt{205}$ For Q5: $x = -3$