MHB Find Zeros of $2x^3+5x^2-28x-15$ with Synthetic Division

  • Thread starter Thread starter karush
  • Start date Start date
AI Thread Summary
The discussion focuses on finding the zeros of the polynomial function g(x) = 2x^3 + 5x^2 - 28x - 15 using synthetic division and the quadratic formula. Synthetic division reveals a factor of (x - 3), leading to a quadratic equation 2x^2 + 11x + 5. The correct application of the quadratic formula results in the roots x = -5 and x = -0.5. The final factorization of g(x) is presented as (x + 10)(x + 0.5)(x - 3). The importance of correctly applying the quadratic formula is emphasized throughout the discussion.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\text{factor}$
$g(x)=2x^3+5x^2-28x-15$
$\text{synthetic division}$
$\begin{array}{c|rr}
& 2 & 5 &-28 &-15 \\
&&6&33 & 15\\ \hline
3&2&11&5&0
\end{array}$
$\text{thus }\\$
$2x^2+11x+5$
$\text{use quadradic formula}$
$\begin{align*}
x=\frac{-(11)\pm\sqrt{(11)^2-4(2)(5)}}{2(1)}
&=\frac{-11\pm\sqrt{81}}{2}
=\frac{-11\pm9}{2}
\end{align*}$
$x=-10,-1 \quad g(x)=(x+10)(x+1)(x-3)$
hopefully
comment?
 
Last edited:
Mathematics news on Phys.org
Check your application of the quadratic formula, or simply factor. :)
 
a=2 not 1

use quadradic formula
$\begin{align*}
x=\frac{-(11)\pm\sqrt{(11)^2-4(2)(5)}}{2(2)}
&=\frac{-11\pm\sqrt{81}}{4}
=\frac{-11\pm9}{4}
\end{align*}$
$\text{then $x=-5,-\frac{1}{2}$
so
$g(x)=(x+10)(x+\frac{1}{2})(x-3)$}$
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Replies
2
Views
1K
Replies
2
Views
1K
Replies
8
Views
1K
Replies
3
Views
1K
Replies
10
Views
3K
Replies
4
Views
1K
Back
Top