Help idk how to do this ASAP please
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When the equation is in the 4th degree, it is called a quartic equation. Solving the roots of a quartic solution is very tedious and lengthy. It could get messy because you would use other variables. If you want to know how, I hope this link could help you: http://mathforum.org/dr.math/faq/faq.cubic.equations.html
I would still give answers but these are determined with the help of tools like scientific calculator and MS Excel.
Part A. This question is very generic. There are many ways of finding the roots. Actually, part b, c and d are all solutions of this section.
Part B. This is how synthetic division works. The divisor is x-3=0 or x=3. Place it on the left side with a corner barrier. Then, place all the coefficients of the quartic equation in order on the right side. This is for the first row. Then, bring down the first coefficient 1 under the horizontal bar. Multiply 1 to the 3 on the left side. The product is 3 which placed under the second coefficient. When you add them, the sum is 0. Multiply this again with 3 on the left side and place this under the next coefficient. The pattern goes on and on.
3| 1 -3 -7 15 18
3 0 -21 -18
----------------------------------
1 0 -7 -6 0
When the last term is 0, it means that there is no remainder. Thus, x-3 is a factor.
Part C. It is quite challenging to factor a quartic equation. As previously mentioned, it can be messy. Just see the link to see the step-by-step procedure. Another alternative to this is by graphical under letter d.
Part D. Just assign values to x, substitute to the equation to get y. Then, plot x against y. The graph would look like that in the picture. The x-intercepts are the roots of the equation. From the picture, we can see that they are x=-1, x=-2 and x=3. The other root is a complex or imaginary.
I would still give answers but these are determined with the help of tools like scientific calculator and MS Excel.
Part A. This question is very generic. There are many ways of finding the roots. Actually, part b, c and d are all solutions of this section.
Part B. This is how synthetic division works. The divisor is x-3=0 or x=3. Place it on the left side with a corner barrier. Then, place all the coefficients of the quartic equation in order on the right side. This is for the first row. Then, bring down the first coefficient 1 under the horizontal bar. Multiply 1 to the 3 on the left side. The product is 3 which placed under the second coefficient. When you add them, the sum is 0. Multiply this again with 3 on the left side and place this under the next coefficient. The pattern goes on and on.
3| 1 -3 -7 15 18
3 0 -21 -18
----------------------------------
1 0 -7 -6 0
When the last term is 0, it means that there is no remainder. Thus, x-3 is a factor.
Part C. It is quite challenging to factor a quartic equation. As previously mentioned, it can be messy. Just see the link to see the step-by-step procedure. Another alternative to this is by graphical under letter d.
Part D. Just assign values to x, substitute to the equation to get y. Then, plot x against y. The graph would look like that in the picture. The x-intercepts are the roots of the equation. From the picture, we can see that they are x=-1, x=-2 and x=3. The other root is a complex or imaginary.