1 If one zero of the polynomial ( a2 + 9 ) x2 + 13x +6a is a reciprocal - Maths - Polynomials

Question

1 If one zero of the polynomial ( a2 + 9 )
x2 + 13x +6a is a reciprocal of the other, find the
values of a Also find the polynomial
2 Factorise p(x) = x3 - 3x2 - 9x - 5
3 If a and b are the zeroes of the quadratic polynomial f (x) =
x2 - p(x+1) - c, show that (a+1) (b+1) = 1 - c
4 The polynomial ax3 + bx2 + x -6 has
(x+2) as a factor and leaves a remainder 4 when divided by (x-2)
Find 'a' and 'b'
5 If the polynomial x4 - 6x3 +
16x2 - 25x +10 is divided by another polynomial
x2 - 2x + k , the remainder comes out to be x+a Find k
and a
6 Find the zeroes of the polynomial f(x) = x3 -
5x2 - 16x + 80 , if its two zeroes are equal in
magnitude but opposite in sign

Ajanta Trivedi · Accepted Answer

the given polynomial be $f (x) = (a^{2} + 9) x^{2} + 13 x + 6 a$

if one zero is the reciprocal of the other.

product of the zeroes = 1.

$\frac{6 a}{a^{2} + 9} = 1 a^{2} + 9 = 6 a a^{2} - 6 a + 9 = 0 a^{2} - 2 * 3 * a + 3^{2} = 0 (a - 3)^{2} = 0$

thus a-3 =0 i.e. a =3

thus the value of a is 3.

2.

$p (x) = x^{3} - 3 x^{2} - 9 x - 5$

at x = -1

$p (- 1) = (- 1)^{3} - 3 * (- 1)^{2} - 9 * (- 1) - 5 = - 1 - 3 + 9 - 5 = - 9 + 9 = 0$

therefore by the factor theorem : $(x + 1)$ is a factor of p(x).

now dividing p(x) by (x+1)

$x + 1 \begin{matrix} x^{2} - 4 x - 5 \\ x^{3} - 3 x^{2} - 9 x - 5 \end{matrix}$

$\underset{̲}{x^{3} + x^{2}} \times - 4 x^{2} - 9 x \underset{̲}{- 4 x^{2} - 4 x} \times - 5 x - 5 \underset{̲}{- 5 x - 5} \times$

thus

$p (x) = (x + 1) (x^{2} - 4 x - 5) = (x + 1) (x^{2} - 5 x + x - 5) = (x + 1) (x - 5) (x + 1) = (x + 1)^{2} (x - 5)$

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