Answer:
(
x
)
=
x
(
x
2
−
x
−
6
)
=
x
(
x
−
3
)
(
x
+
2
)
=
0
x
=
0
,
x
=
3
,
x
=
−
2
Explanation:
To find the zeros of the polynomial above first factor out the common factor then factor the remaining polynomial and set it equal to zero. Finally solve for x.
Step-by-step explanation:
Answer:
695
Step-by-step explanation:
(x-5)(x+3)=0
x^2+3x-5x-15=0
(x^2+3x) (-5x-15)
x(x+3) -5(x+3)
x-5=0 x+3=0
x=5 OR -3
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B2%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D2%5C%5C%20k%3D-1%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-2%29%5E2-1%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D0%5C%5C%20x%3D5%20%5Cend%7Bcases%7D%5Cimplies%200%3Da%285-2%29%5E2-1%5Cimplies%201%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B9%7D%3Da%5Cqquad%20therefore%5Cqquad%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%7D)
now, let's expand the squared term to get the standard form of the quadratic.
The answer is x= -17
hope this helps
