Check weather(-2x-5) is a factor of the polynomial p(x)=3x^4+5x^3-2x^2-4
Answers
Step-by-step explanation:
Answer:
(-2x-5) \ is \ not \ a \ factor \ of \ the \ polynomial \ p(x)=3x^4+5x^3-2x^2-4(−2x−5) is not a factor of the polynomial p(x)=3x
4
+5x
3
−2x
2
−4
Step-by-step explanation:
\begin{gathered}\large \text{Given $p(x)=3x^4+5x^3-2x^2-4 \ and \ g(x)=-2x-5$}\\\\\\\\large \text{zeroes of g(x)=-2x-5=0}\\\\\\\\\large \text{$x=\dfrac{-5}{2}$}\\\\\\\large \text{putting g(x) value in p(x)}\\\\\\\\\\\large \text{$p(\dfrac{-5}{2})=3(\dfrac{-5}{2})^4+5(\dfrac{-5}{2})^3-2(\dfrac{-5}{2})^2-4 $}\\\\\\\\\\\large \text{$p(\dfrac{-5}{2})=3(\dfrac{625}{16})+5(\dfrac{-125}{8})-2(\dfrac{25}{4})-4 $}\\\\\\\\\\\\\end{gathered}
Given p(x)=3x
4
+5x
3
−2x
2
−4 and g(x)=−2x−5
largezeroes of g(x)=-2x-5=0
x=
2
−5
putting g(x) value in p(x)
p(
2
−5
)=3(
2
−5
)
4
+5(
2
−5
)
3
−2(
2
−5
)
2
−4
p(
2
−5
)=3(
16
625
)+5(
8
−125
)−2(
4
25
)−4
\begin{gathered}\large \text{$p(\dfrac{-5}{2})=(\dfrac{1875}{16})-(\dfrac{625}{8})-(\dfrac{50}{4})-4 $}\\\\\\\\\\\large \text{$p(\dfrac{-5}{2})=(\dfrac{1875-1250-200-64}{16})$}\\\\\\\\\\\large \text{$p(\dfrac{-5}{2})=\dfrac{1875-1514}{16}$}\\\\\\\\\\\large \text{$p(\dfrac{-5}{2})=\dfrac{361}{16}$}\\\\\\\\\\\large \text{Since remainder does not come 0 }\\\\\\\\\\\large \text{Therefore g(x) is not a factor of (p)x}\end{gathered}
p(
2
−5
)=(
16
1875
)−(
8
625
)−(
4
50
)−4
p(
2
−5
)=(
16
1875−1250−200−64
)
p(
2
−5
)=
16
1875−1514
p( 2−5 )= 16
361
Since remainder does not come 0
Therefore g(x) is not a factor of (p)x