Simplify (x3 + 3x2 + 5x – 4) – (3x3 – 8x2 – 5x + 6)
The first thing I have to do is take that "minus" sign through the parentheses containing the second polynomial. Some students find it helpful to put a "1" in front of the parentheses, to help them keep track of the minus sign.
Here's what the subtraction looks like, when working horizontally:
(x3 + 3x2 + 5x – 4) – (3x3 – 8x2 – 5x + 6)
(x3 + 3x2 + 5x – 4) – 1(3x3 – 8x2 – 5x + 6)
(x3 + 3x2 + 5x – 4) – 1(3x3) – 1 (–8x2) – 1(–5x) – 1(6)
x3 + 3x2 + 5x – 4 – 3x3 + 8x2 + 5x – 6
x3 – 3x3 + 3x2 + 8x2 + 5x + 5x – 4 – 6
–2x3 + 11x2 + 10x –10
And here's what the subtraction looks like, when going vertically:
x
3
−(3x
3
+3x
2
−8x
2
+5x
−5x
−4
+6)
In the horizontal addition (above), you may have noticed that running the negative through the parentheses changed the sign on each and every term inside those parentheses. The shortcut when working vertically is to not bother writing in the subtaction sign or the parentheses; instead, write the second polynomial in the second row, and then just flip all the signs in that row, "plus" to "minus" and "minus" to "plus".
\
x
3
–3x
3
−2x
3
+3x
2
+8x
2
+11x
2
+5x
+5x
+10x
−4
–6
−10
Either way, I get the answer:
–2x3 + 11x2 + 10x – 10
Answer:hi
Step-by-step explanation:hello
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.](https://tex.z-dn.net/?f=5m%5E2e%5E%7Bmt%7D%2B3me%5E%7Bmt%7D-2e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%285m%5E2%2B3y-2%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B3m-2%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%205m%5E2%2B5m-2m-2%3D0%5C%5C%5C%5C%5CRightarrow%205m%28m%2B1%29-2%28m%2B1%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B1%29%285m-1%29%3D0%5C%5C%5C%5C%5CRightarrow%20m%2B1%3D0%2C~~~~~5m-1%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-1%2C~%5Cdfrac%7B1%7D%7B5%7D.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
![y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D2.8%5C%5C%5C%5C%5CRightarrow%20-A%2B%5Cdfrac%7BB%7D%7B5%7D%3D2.8%5C%5C%5C%5C%5CRightarrow%20-5A%2BB%3D14%5C%5C%5C%5C%5CRightarrow%20-5A-A%3D14~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7BUisng%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-6A%3D14%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B14%7D%7B6%7D%5C%5C%5C%5C%5CRightarrow%20A%3D-%5Cdfrac%7B7%7D%7B3%7D.)
From equation (ii), we get
Thus, the required solution is
The first option is correct
