Question

Find the value of a if (x-a) is a factor of the polynomial ^6-ax^5+x^4-ax^3+3x-a+2 if both (x-2) and (x-1/2) are the factors of px^2+5x+r ,show that p=r

Answer 1

these are the 2 different questions.
for first put x= a into the polynomial and it will becomes 0
${x}^{6} - a {x}^{5} + {x}^{4} - a {x}^{3} + 3x - a + 2 \\ {a}^{6 } - {a}^{6} + {a}^{4} - {a}^{4} + 3a - a + 2 = 0 \\ 2a + 2 = 0 \\ a = - 1$
for second question put the value x=2 and x= 1/2 and equate both equations to zero
$p {x}^{2} + 5x + r \\ put \: x = 2 \\ 4p + 10 + r = 0 \\ 4p + r = - 10 \: \: \: \: ...eq \: 1$
now put x= 1/2
$\frac{p}{4} + \frac{5}{2} + r = 0 \\ p + 10 + 4r = 0 \\ p + 4r = - 10 \: \: \: ...eq \: 2$
multiply eq 1 by 4 and subtract both of them
$16p + 4r = - 40 \\ p + 4r = - 10 \\ 15p = - 30 \: \: \: and \: p = - \frac{30}{15} = - 2$
put the value of p in any of the equation
$- 2 + 4r = - 10 \\ 4r = - 10 + 2 \\ r = \frac{ - 8}{4} \\ r = - 2$
so from the value of p and r we can say that p=r