Question

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If both (x - 2) and (x - 2) are factors of px? +5x+r, prove that p = r.​

Answer 1

The given polynomial is Q(x)=px²+5x+r

It is also given that (x−2) and (x-1/2) are the factors of Q(x) which means that Q(2)=0 and Q(1/2)=0.

Let us first substitute Q(2)=0 in Q(x)=px²+5x+r as shown below :

Q(x)=px²+5x+r

=> Q(2)=p(2)²+5(2)+r

0=4p+10+r

0=4p+10+r

4p+r=−10.........(1)

Now, substitute Q(1/2)=0 in Q(x)=px²+5x+r as shown below :

Q(x)=px²+5x+r

=> Q(1/2)=p(1/2)²+5(1/2)+r

0=p/4+5/2+r

0=p+5(2)+4(r)/4

0=4r+10+p/4

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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