Question

if both (x-2) and (x-1/2) are factorise of px2+5x+r.show that p=r​

Answer 1

Explanation:

Let the polynomial be f(x).

If (x - 2) and (x- 1/2) are factors of f(x), this means that f(2) = f(1/2) =0.

$f(2)=p(2)^2+5(2)+r$

==> $f(2) = 4p+r+ 10$

==> $4p+r=-10$ ------ i. [∵ f(2) = 0]

$f(\frac{1}{2})=p(\frac{1}{2})^2+5(\frac{1}{2})+r$

==> $f(\frac{1}{2})=\frac{1}{4}p+r+\frac{5}{2}$

==> $\frac{1}{4}p+r=-\frac{5}{2}$ ------- ii. [∵f(1/2)=0]

When we subtract equation ii. from equation i,

$4p + r - \frac{1}{4}p-r=-10-(-\frac{5}{2})$

==> $4p -\frac{1}{4}p=-10+\frac{5}{2}$

==> $\frac{16-1}{4}p=\frac{-20+5}{2}$

==> $\frac{15}{4}p=\frac{-15}{2}$

==> $p = \frac{-15}{2} \times\frac {4}{15}=\frac{-1}{1}\times \frac{2}{1}$

∴ $p=-2$

Hence, by substituting the value of p in equation i,

$4(-2)+r=-10$

==> $r-8=-10$

==> $r=-10+8$

∴ $r=-2$

Hence, we have shown that p = r = -2.

Answer: p = r = -2

I hope this helps. :D