If both x-2 and x-1/4 factors ofpx + 5x + r, show that p =r
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Answer:
given polynomial is Q(x)=px
2
+5x+r
It is also given that (x−2) and (x−
2
1
) are the factors of Q(x) which means that Q(2)=0 and Q(
2
1
)=0.
Let us first substitute Q(2)=0 in Q(x)=px
2
+5x+r as shown below:
Q(x)=px
2
+5x+r
⇒Q(2)=p(2)
2
+(5×2)+r
⇒0=(p×4)+10+r
⇒0=4p+10+r
⇒4p+r=−10.........(1)
Now, substitute Q(
2
1
)=0 in Q(x)=px
2
+5x+r as shown below:
Q(x)=px
2
+5x+r
⇒Q(
2
1
)=p(
2
1
)
2
+(5×
2
1
)+r
⇒0=
4
p
+
2
5
+r
⇒0=
4
p+(5×2)+(r×4)
⇒0=
4
p+10+4r
⇒0×4=p+10+4r
⇒0=p+10+4r
⇒p+4r=−10.........(2)
Now subtracting the equations (1) and (2), we get
(4p−p)+(r−4r)=−10−(−10)
⇒3p−3r=−10+10
⇒3p−3r=0
⇒3p=3r
⇒p=r
Hence, p=r is proved.
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