SOLUTION: 1. What is the determinant of the matrix [-1 2][1 3]. The [1 3] is directly under the [-1 2] in the problem. I have no idea where to begin to solve this problem. 2. The below ta

Question 46992This question is from textbook CLEP Official Study Guide
: 1. What is the determinant of the matrix [-1 2][1 3]. The [1 3] is directly under the [-1 2] in the problem. I have no idea where to begin to solve this problem.
2. The below table gives some of the values of a 5th degree polynominal p(x). Based on the values shown, what it the minimum number of real roots fot the equation p(x)=0
x 0 1 2 3 4 5 6 7
p(x) -30 22 110 150 34 -130 222 2,350
I have no idea how to begin to solve this problem. This question is from textbook CLEP Official Study Guide

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
1. What is the determinant of the matrix [-1 2][1 3]. The [1 3] is directly under the [-1 2] in the problem. I have no idea where to begin to solve this problem.
|-1,2|=(-1)(3)-(2)(1)=-3-2=-5
| 1,3|
IN GENERAL
|A,B|
|C,D|=A*D-B*C
2. The below table gives some of the values of a 5th degree polynominal p(x). Based on the values shown, what it the minimum number of real roots fot the equation p(x)=0
x 0 1 2 3 4 5 6 7
p(x) -30 22 110 150 34 -130 222 2,350
I have no idea how to begin to solve this problem.
REAL ROOTS ARE THOSE VALUES OF X AT WHICH P(X) BECOMES ZERO.
WE FIND THAT AT
X=0.....P(X)=-30..AND
AT X=1...P(X)=+22.....SO P(X) WOULD BE ZERO BETWEN X=0 AND X=1..THIS IS ONE REAL ROOT
SIMILARLY...AT
X=4.....P(X)=34
AT X=5....P(X)=-130.....SO P(X) WOULD BE ZERO BETWEEN X=4 AND X=5...THI9S IS ANOTHER REAL ROOT
SIMILARLY...AT X=5.....P(X)=-130
AT X=6............P(X)=222..HENCE P(X)WOULD BE ZERO BETWEEN X=5 AND X=6...THIS IS ANOTHER REAL ROOT.
HENCE THERE ARE 3 REAL ROOTS IN ALL
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