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alexandr1967 [171]
3 years ago
5

Divide 3x^2-x-2 by x+4

Mathematics
2 answers:
jasenka [17]3 years ago
7 0
The answer would be 3x^2-13+50/x+4

Explanation:
You can use synthetic division to solve this.
How you do synthetic division you have to take the number on x+4 and change its sign

Which is -4

Then you take A B and C (in this case A B and C is 3, -1, and -2)

To do synthetic division you bring the 3 down and multiply it by -4 and add the answer to the -1

Then you have 3-13 which you can change into 3x-13. For this specific problem it has a remainder.

You find the remainder by going back to your -4 and multiply it by the -13 (which is 52 because a negative times a negative equals a positive) and add the 52 to -2 which is 50.

So your final answer is 3x-13+50/x+4

I’m sorry if this was confusing written out like this. If you want to learn how to do synthetic division search it up.

Hope this helps
drek231 [11]3 years ago
5 0

Answer: 3x-13+50/x+4

Step-by-step explanation:

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Rewrite the following system of linear equations in matrix equation form and in vector equation form. Solve the system.
alisha [4.7K]

Answer:

The set of solutions is \{\left[\begin{array}{c}a\\b\\x\\y\\z\end{array}\right] = \left[\begin{array}{c}-26+503y+543z\\-37+655y+724z\\-4+80y+90z\\y\\z\end{array}\right] : \text{y, z are real numbers}\}

Step-by-step explanation:

The matrix associated to the problem is A=\left[\begin{array}{ccccc}1&-1&2&-8&1\\2&-1&-4&1&-2\\-4&1&4&-3&-1\end{array}\right] and the vector of independent terms is (3,1,-1)^t. Then the matrix equation form of the system is Ax=b.

The vector equation form is a\left[\begin{array}{c}1\\2\\-4\end{array}\right]+b\left[\begin{array}{c}-1\\-1\\1\end{array}\right] + x\left[\begin{array}{c}2\\-4\\4\end{array}\right]+y\left[\begin{array}{c}-8\\1\\-3\end{array}\right] + z\left[\begin{array}{c}1\\-2\\-1\end{array}\right]=\left[\begin{array}{c}3\\1\\-1\end{array}\right].

Now we solve the system.

The aumented matrix of the system is \left[\begin{array}{cccccc}1&-1&2&-8&1&3\\2&-1&-4&1&-2&1\\-4&1&4&-3&-1&-1\end{array}\right].

Applying rows operations we obtain a echelon form of the matrix, that is \left[\begin{array}{cccccc}1&-1&2&-8&1&3\\0&1&-8&-15&-4&-5\\0&0&1&-80&-9&-4\end{array}\right]

Now we solve for the unknown variables:

  • x-80y-90z=-4 then x=-4+80y+90z
  • b-8x-15y-4z=-5, b-8(-4+80y+90z)-15y-4z=-5 then b=-37+655y+724z.
  • a-b+2x-8y+z=3, a-(-37+655y+724z)+2(-4+80y+90z)-8y+z=3, then a=-26+503y+543z

Since the system has two free variables then has infinite solutions.

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