Without actual division, show that (2x^2 − 7x−15) is a factor of (2x^3– 9x^2 –8x + 15)
Answers
Proof :
Let, p(x)=x
4
+2x
3
−2x
2
+2x−3
And, g(x)=x
2
+2x−3
Then, g(x)=x
2
+2x−3
=x
2
+3x−x−3
=x(x+3)−(x+3)
=(x+3)(x−1)
Now, we check if g(x) is a factor of p(x) by using factor theorem.
∴ (x+3) and (x−1) divides p(x) ifp(−3) and p(1)=0
So,
p(−3)=(−3)
4
+2(−3)
3
−2(−3)
2
+2(−3)−3
=81−54−18−6−3=0
and,
p(1)=(1)
4
+2(1)
3
−2(1)
2
+2(1)−3
=1+2−2+2−3=0
Hence, p(x) is divisible by (x+3) and (x−1)
⇒p(x) is divisible by (x+3)(x−1)
⇒p(x) is divisible by g(x)
Hence proved...
please mark me as brainliest
Answer:
x
4
+2x
3
−2x
2
+2x−3 is exactly divisible by x
2
+2x−3
Proof :
Let, p(x)=x
4
+2x
3
−2x
2
+2x−3
And, g(x)=x
2
+2x−3
Then, g(x)=x
2
+2x−3
=x
2
+3x−x−3
=x(x+3)−(x+3)
=(x+3)(x−1)
Now, we check if g(x) is a factor of p(x) by using factor theorem.
∴ (x+3) and (x−1) divides p(x) ifp(−3) and p(1)=0
So,
p(−3)=(−3)
4
+2(−3)
3
−2(−3)
2
+2(−3)−3
=81−54−18−6−3=0
and,
p(1)=(1)
4
+2(1)
3
−2(1)
2
+2(1)−3
=1+2−2+2−3=0
Hence, p(x) is divisible by (x+3) and (x−1)
⇒p(x) is divisible by (x+3)(x−1)
⇒p(x) is divisible by g(x)
Hence proved.