Question

if (x-2) and (x-1/2) are the factors of the polynomial qx2+5x+r prove that q=r

Answer 1

h(x) = (x-2) and g(x) = (x- 1/2) are the factors of p(x) = qx²+5x+r

As both are factors of the polynomial, 0 will be the outcome.

h(x) = x-2
h(x) = 0
=> x-2 = 0 => x = 2
x-2 is a factor. so,
p(2) = 0
q(2)²+5(2)+r = 0
4q+ 10 + r = 0
r = -(4q+10) .... (1)

g(x) = x- 1/2
g(x) = 0
=> x- 1/2 = 0 => x = 1/2
x-1/2 is a factor.
p(1/2) = 0
p(1/2) q(1/2)²+5(1/2)+r = 0
q/4 + 5/2 + r =0
r = - (q+10/4) ..... (2)

p(2) = p(1/2)
equation (1) = equation (2)
-(4q+10)= -(q+10/4)
4q+10 = q+10/4
4 (4q+10) = q+10
16q+40 = q + 10
15q= -30
q= -2

Finding r.

r = -(4q+10)
r = - (-8+10)
r = -2

∴ q = r = -2 (hence proved)

Answer 2

Answer:

p(x) = qx²+5x +r

x-2= 0

x=2

p(2)= q(2)²+ 5(2)+ r

=4q+10+r=0

=r= -(4q+10) ............(1)

p(1/2)=q(1/2)²+5(1/2)+r

=q/4+5/2+r=0

=q+10/4 +r =0

= r= - (q+10/4)..........(II)

eq. (l) =eq.(ll)

-4q+10= -(q+10/4)

4q+10= q+10/4

16q+40=q+10

16q-q=10-40

15q= -30

q= -2

from eq. (l)

r= -(4q+10)

r= -(4× -2 +10)

r= -( -8+10)

r= -2

therefore, q=r= -2

proved