MCQ : Polynomials - 1

x=5

putting x in equation

p(x)=x²-kx-15=0

=(5)²-k(5)-15=0

=25-5k-15=0

=10-5k=0

=-5k=-10

=k=-10/-5

=k=2

Value of k is 2

Zero of p(x)
Let p(x) = ax + b
Put x = k
p(k) = ak + b = 0
∴ k is zero of p(x)

Given, p(x) = 2x+5
For zero of the polynomial, put p(x) = 0
∴ 2x + 5 = 0
⇒ -5/2

Hence, zero of the polynomial p(x) is -5/2. 

 (k - l)x² + kx +1

One zero is - 3, so it must satisfy the equation and make it zero
∴ (k - 1) (-3)² + k(-3) + 1 - 0
⇒ 9k - 9 - 3k + 1 = 0
⇒ 6k - 8 = 0
⇒ 

Let p(x) = x² + ax + b.
Put a = 0, then,  p(x) = x² + b = 0
⇒    x² = -b
⇒  x =  ±√-b
[∴b < 0]
Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = 0 and the constant term is negative i.e., b< 0.
Alternate Method
Let   f(x) = x² + ax+ b
and by given condition the zeroes area and – α.
Sum of the zeroes = α- α = a
= a = 0
f(x) = x² + b, which cannot be linear,
and product of zeroes = α .(- α) = b
⇒ -α² = b
which is possible when, b < 0.
Hence, it has no linear term and the constant term is negative.

 x² + (a + 1)x + b
∵  x = 2 is a zero and x = - 3 is another zero 
∴ (2)² + (a + 1)² + b = 0
and (-3)² + (a + 1) (-3) + b = 0
⇒ 4 + 2a + 2 + b = 0 and 9 - 3a - 3 + b = 0
⇒ 2a + b = -6 ...(i) and -3a + b = -6 ..(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = -6. 

 Let p (x) = ax² + bx + c be the required polynomial whose zeroes are -2 and 5.

∴ Sum of zeroes = -b/a
⇒  ...(i)
and product of zeroes = c/a
⇒  ...(ii)
From Eqs. (i) and (ii)
a = 1, b = -3 and c = -10
∴ p(x) = ax² + bx + c = 1.x² - 3x - 10
= x² - 3x - 10
But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change.
∴ p(x) = kx² - 3kx - 10k [where, k is a real number]
⇒  [where, k is a non-zero real number]

Hence, the required number of polynomials are infinite i.e., more than 3.

For any quadratic polynomial ax² + bx + c, a≠0, the graph of the Corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like u or open downwards like  ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. 
Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (a) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial. 

Let the roots be α and 1/α. Then α(1/α) = m/5 or 1 = m/5 or m = 5

Sum of zeroes = -b/a