Solve cubic equations by prime factorization! ? How to solve 4th and 5th order equations! !

Hello. I'm Tenmei Watanabe from Seito Medical School Clear File. This time I'd like to think about cubic expressions.

Now, suppose we have the expression a^3+103(a^2)+3431a+36593=0. Quickly, this formula is (a+23)(a+37)(a+43)=0, and the answer is a = -23, -37, -43.

Here, a^3 is called the first term, 103(a^2) is the second term, 3431a is the third term, and 36593 is the fourth term.

You can see that the fourth term, 36593, is exactly (-23)x(-37)x(-43) multiplied by (-1). In other words, if you can decompose 36593 into prime factors, you can find the value of a.

Now, let's try dividing this cubic function using long division. The second term is a^2 multiplied by 103. Since this function is a cubic expression, dividing 103 by 3 gives us 34.333..., so we round it down to 34.

Let's take this 34 as a+34 and divide a^3+103(a^2)+3431a+36593. First, let's subtract a^3+103(a^2) by (a+34)x(a^2). a^3 +103(a^2) – a^3 – 34(a^2) = 69(a^2).

The third term, 3431a, and the fourth term, 36593, are the remainder, so adding 69(a^2) gives 69(a^2)+3431a+36593. If we further divide this remainder expression by a+34, we can see that we need to subtract (a+34)x69a=(a^2)+2346a.

The remainder will be 1085a+36593. Furthermore, if we subtract (a+34)x1085 from the remainder, the remainder becomes -297.

The final result is a^3+103(a^2)+3431a+36593=(a+34)(a^2+69a+1085)-297.

Since -297=(-1)x3x3x3x11, this is a combination problem, and we can see that a+34+3=a+37, a+34+(-1)x11=a+23, a+34+3×3=a+43, which can be decomposed into prime factors.

Using this method, you should be able to solve both 4th and 5th order equations. that's all. See you soon.