Here's the question you clicked on:
LovinBachata
Find The Four Real Zeros Of The Polynomial. f(x)= x^4+6x^3-15x^2-152x-240
Descartes rule of signs tells us that there is only one positive root. Factoring theorem tells us that the rational roots would be factors 240 (which has many). To guess the positive root, seeing that the first two terms are positive, with small coefficients, and the remaining (negative) have large coefficients, we can try large factors, such as 4 (4^4=256), or 5 (5^4=625) by evaluating f(4) or f(5). Once we find the first one, the rest are negative roots of the remaining cubic.
so the anser would be -4,-4,-3,5 ?
Yes, how did you find them?
i first used synthetic division using the number 4 and -4 and kept dividing until i had four solutions
Yep, that's great! Not everyone is familiar with synthetic division. How did you get to start with 4 and -4?
Synthetic division is easier for me to use and because you gave me those numbers
I hope you'd be able to find those starter numbers in an exam using Descartes rule of signs and the factor theorem!
I Hope So Tooo! Thank You Soo Much
You synthetic division capability is a great asset in solving equations. Another way is to evaluate f(x) by nesting, for example, for f(5), you can almost do it by inspection (mentally): f(5)=(((5+6)*5-15)*5-152)*5-240 =((55-15)*5-152)*5-240 =(200-152)*5-240 =48*5-240 =0
but how do you know with what numbers to start with?
Assuming you have a starter number of 5, then evaluating using nesting (as shown above) is fast, accurate and less error prone... with a little practice. The starter number remain guesses, with he help of Descartes Theorem and Factor theorem. Are you already familiar with these two?
im only familiar with the factor theorem but i dont remember Descartes theroem do you usuallly look ath the last numbeer?
If the leading coefficient is 1, then only the last number counts. If the leading coefficient id different from one, then the possible factors are of the form (ax+b) where a is a factor of the leading coefficient, and b is a factor of the last number (constant term). Descartes rule tells us that the maximum number of positive roots is equal to the number of sign changes. In the above case, sign only changed once from + to -, so there is only one positive root, and a maximum of three negative roots. Maximum means 2,4,6... can be replaced by complex roots, since complex roots always come in pairs. For example, in our case, there is one positive root (cannot be replaced by two complex), and one or three real negative roots. Since the question says 4 real roots, we can count on 3 negative (real) roots.
* replace (ax+b) by \( (ax\pm b)\)
so the fact that there is one postive sign means that the solutions would have one positive solution?
It's the \( change \) in sign that counts. x^2-2x+1=0 has two positive roots, but x^2-x-2=0 has only one.