Question

If both (x-2) and (x-1/2) are factors of px^2+5x+r.Show that p=r.​

Answer 1

Given, f(x) = px^2+5x+r  and factors are x-2, x-1/2

Substitute x = 2 in place of equation, we get

= p*2^2+5*2+r=0  

=  4p + 10 + r = 0   ------------  (i)

Substitute x = 1/2 in place of equation.  

 p/4 + 5/2 + r = 0

 p + 10 + 4r = 0  -------------------  (ii)

On solving (i),(ii) we get

4p+r=-10 and p+4r+10=0

4p+r=p+4r

3p=3r   

p = r.

Hope this helps!

PLZ MARK AS BRAINLIST ANSWER

Answer 2

Answer:

Step-by-step explanation:

Let f(x) = px2 + 5x + r

By Factor Theorem, we know that,

If p(x) is a polynomial and a is any real number, then g(x) = (x– a) is a factor of p(x), if p(a) = 0 and vice versa.

So, if (x – 2) is a factor of f(x)

⇒ f(2) = 0

⇒ p(2)2 + 5(2) + r = 0

⇒ 4p + r = –10 -------- (A)

Also as is also a factor,

⇒ p + 4r = –10 ----- (B)

Subtract B from A to get,

4p + r - (p + 4r)= –10 - (-10)

4p + r - p - 4r = -10 + 10

3p - 3r = 03p = 3rp = r