+26396 Find 
if the expansion of the product of 
and 
has no 
term.
(x3−4x2+2x−5)∗(x2+tx−7)=tx4−4tx3+2tx2−5tx+x5−4x4−5x3+23x2−14x+35
Set 2tx2+23x2=0 than the product has no x2, t must be a constant! 2tx2+23x2=02tx2=−23x2|:2x2t=−232=−11.5
t=−11.5(x3−4x2+2x−5)∗(x2+(−11.5)x−7)=x5−15.5x4+41x3+43.5x+35
There is no more x2
+26396 ----------------------------------------------------------
The polynomial 
has degree 3. If 
, 
, 
, and 
, then what are the x-intercepts of the graph of ?
The polynomial f(x) of degree 3 is f(x)=ax3+bx2+cx+d
I. We need a, b, c and d :
\begin{array}{r|r|lrclrclccl} \hline x & y & &f(x)& =& ax^3+bx^2+cx+d & && &\textcolor[rgb]{1,0,0}{d}&\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{0} \ \hline -1 & 15& (1) & 15 &=& a(-1)^3+b(-1)^2+c(-1) +d & 15&=& -a+b-c+d & 15&=&-a+b-c\ 0 & 0 & (2) & 0 &=& a(0)^3+b(0)^2+c(0) +d & \textcolor[rgb]{1,0,0}{0} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{d} & -5&=&a+b+c\ 1 & -5 & (3) & -5 &=& a(1)^3+b(1)^2+c(1) +d & -5 &=& a+b+c+d & \ 2 & 12 & (4) & 12 &=& a(2)^3+b(2)^2+c(2) +d & 12 &=& 8a+4b+2c+d & 12 &=& 8a+4b+2c\ \hline \end{array}
d=0:
(1) -a + b - c = 15
(2) a + b + c = -5
(4) 8a+4b+2c = 12 | :2 ⇒ (4) 4a + 2b + c = 6
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(1)+(2): 2b = 10 ⇒ b=5
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b=5:
(1) a + c = -10
(2) a + c = -10
(4) 4a + c = -4
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(4)-(2): 3a = -4 -(-10) = 6 ⇒ 3a = -4+10 ⇒ 3a = 6 => a=2
(1) 2 + c = -10 ⇒ c=−12
The polynomial f(x) of degree 3 is f(x)=2x3+5x2−12x+0
II. x-intercepts of the graph of f(x)?
2x3+5x2−12x=0x⏟=0∗(2x2+5x−12)⏟=0=0x1=02x2+5x−12=0|ax2+bx+c=0=>x=−b±√b2−4ac2ax2,3=−5±√25−4∗2∗(−12)4x2,3=−5±√1214x2,3=−5±\114x2=−5+114=64=1.5⇒x2=1.5x3=−5−114=−164=−4⇒x3=−4
The x-intercepts are: -4, 0 and 1.5
