pa tulong po math grade 10 2nd quarter
Answer:
To solve these equations, we need to find the values of \(x\) that satisfy each equation. Let’s solve them one by one:
1) \(x^{2}-12-x=0\) can be rearranged as \(x^{2}-x-12=0\). This is a quadratic equation. Factoring or using the quadratic formula, we can find that the solutions are \(x = -3\) and \(x = 4\).
2) \(x^{2}-16=0\) can be rewritten as \(x^{2}-4^{2}=0\), which is a difference of squares. This equation can be factored as \((x-4)(x+4)=0\). The solutions are \(x = -4\) and \(x = 4\).
3) \(2x^{2}+x^{3}-5x-6=0\) is a cubic equation. Factoring or using other methods, we can find that one of the solutions is \(x = -2\). We can then divide the equation by \(x+2\) to get \(x^{2}-x-3=0\), which is a quadratic equation. We can factor this as \((x-3)(x+1)=0\). So the solutions to the original equation are \(x = -2\), \(x = -1\), and \(x = 3\).
4) \(x^{3}+6x^{2}+11x+6=0\) is a cubic equation. By using synthetic division or other methods, we can find that one of the solutions is \(x = -1\). Dividing the equation by \(x+1\), we get \(x^{2}+5x+6=0\). This quadratic equation can be factored as \((x+2)(x+3)=0\). So the solutions to the original equation are \(x = -1\), \(x = -2\), and \(x = -3\).
5) \(x^{4}+2x^{3}-3x^{2}-8x-4=0\) is a quartic equation. By using factoring or other methods, we can find that \(x = -1\) is a solution. Dividing the equation by \(x+1\), we get \(x^{3}+x^{2}-4x-4=0\). By synthetic division or other methods, we find that \(x = -1\) and \(x = 2\) are solutions. Dividing the equation by \(x-2\), we get \(x^{2}+3x+2=0\), which can be factored as \((x+1)(x+2)=0\). So the solutions to the original equation are \(x = -1\), \(x = -2\), and \(x = 2\).
6) \(2x^{2}-3x-2=0\) is a quadratic equation. We can factor this as \((2x+1)(x-2)=0\). So the solutions are \(x = -\frac{1}{2}\) and \(x = 2\).
7) \(-17x^{2}+18x+3x^{3}+8=0\) can be rearranged as \(3x^{3}-17x^{2}+18x+8=0\). By using synthetic division or other methods, we find that \(x = -\frac{2}{3}\) is a solution. Dividing the equation by \(x+\frac{2}{3}\), we get \(3x^{2}-20x+12=0\). This quadratic equation can be factored as \((3x-2)(x-6)=0\). So the solutions to the original equation are \(x = -\frac{2}{3}\), \(x = \frac{2}{3}\), and \(x = 6\).
8) \(2x^{3}-18x=0\) can be factored as \(2x(x^{2}-9)=0\). This equation has three solutions: \(x = 0\), \(x = -3\), and \(x = 3\).
9) \(x^{4}-7x^{2}+6x=0\) can be factored as \(x(x^{3}-7x+6)=0\). This equation has two solutions: \(x = 0\) and \(x = 1\).
10) \(x^{5}-3x^{4}-11x^{3}+27x^{2}+10x-24=0\) is a quintic equation. By using synthetic division or other methods, we find that \(x = 1\) is a solution. Dividing the equation by \(x-1\), we get \(x^{
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