Polynomials – Notes, MCQs, Quiz & Worksheet

A quadratic polynomial with zeros α and β can be written as with k is a non zero constant :

k(x² – (α + β)x + αβ)

Where α and β are the roots of the polynomial.

Here it is given that α = 0 and β = √5

So, the polynomial will be:

k(x² – (0 + √5)x + 0(√5))

k(x² – √5x)

Let k = 1 for simplicity the equation is   x² – √5x

Let, f(x) = x² – 6x + y

And it is given that

3α + 2β = 20———————(i)

As α and β are zeroes of the quadratic polynomial x² – 6x + y 

So   α + β = -b/a = -(-6)/1

=> α + β = 6 ———————(ii)

And,

αβ = y———————(iii)

Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),

=> α = 20 – 12 = 8

Now, substitute this value in equation (ii),

=> β = 6 – 8 = -2

Substitute the values of α and β in equation (iii) to get the value of y, such as;

y = αβ = (8)(-2) = -16

Hence value of y is -16

As 2 is the zero of the polynomial.

We know that if α is a zero of the polynomial p(x), then p(α) = 0

Put x = 2 in  equation x² + 3x + p

⇒ 2² + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

Hence P = -10 is the answer

If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, find the value of a.

For a quadratic polynomial ax² + bx + c, if one zero is the reciprocal of the other, then the product of zeros is 1.

Therefore, c/a = 1, which means 4k / (k² + 4) = 1. Solving this gives k² - 4k + 4 = 0, so (k-2)² = 0, hence k=2.

The zeros of a polynomial are the x-coordinates where its graph intersects the x-axis.

The given graph intersects the x-axis at three distinct points, indicating three zeros.

The polynomial is f(x) = x² - px - p - c. Here, a=1, b=-p, c=-(p+c).

So, alpha + beta = p and alpha * beta = -(p+c). (alpha + 1)(beta + 1) = alpha*beta + alpha + beta + 1 = -(p+c) + p + 1 = 1 - c.

The maximum number of real zeros a polynomial can have is equal to its degree.

A cubic polynomial has a degree of 3, so it can have at most 3 real zeros.

For a cubic polynomial ax³ + bx² + cx + d, the product of zeros (alpha*beta*gamma) = -d/a.

Here, -d/a = -(-6)/1 = 6. Given that alpha*beta = 2, then 2 * gamma = 6, which implies gamma = 3.

If -3 is a zero, then substituting x=-3 into the polynomial should result in 0.

(k-1)(-3)² + k(-3) + 1 = 0 => 9(k-1) - 3k + 1 = 0 => 9k - 9 - 3k + 1 = 0 => 6k - 8 = 0 => 6k = 8 => k = 8/6 = 4/3.

For a quadratic polynomial with zeros alpha and beta, the polynomial is k(x² - (alpha+beta)x + alpha*beta).

Here, alpha+beta = -2+5 = 3, and alpha*beta = -2*5 = -10. So, the polynomial is x² - 3x - 10 (taking k=1).

From the polynomial, alpha + beta = -6 and alpha * beta = 2.

(1/alpha + 1/beta) = (alpha + beta) / (alpha * beta) = -6 / 2 = -3.

Sum of zeros = 2+3 = 5. Product of zeros = 2*3 = 6.

From the polynomial, sum of zeros = -(-2k)/3 = 2k/3. So, 2k/3 = 5 => k = 15/2. Product of zeros = 2m/3. So, 2m/3 = 6 => m = 9.

Sum of zeros = -2/k. Product of zeros = 3k/k = 3.

Given sum = product, so -2/k = 3. This implies k = -2/3.

alpha + beta = 8 and alpha * beta = k.

We know alpha² + beta² = (alpha + beta)² - 2alpha*beta. So, 40 = (8)² - 2k => 40 = 64 - 2k => 2k = 24 => k = 12.

Let the zeros be alpha, beta, gamma. We are given alpha = 0 and beta = 0.

The sum of the zeros is alpha + beta + gamma = -b/a. Substituting the given zeros, 0 + 0 + gamma = -b/a, so gamma = -b/a.

Using polynomial long division, dividing x⁴ + 2x³ + 8x² + 12x + 18 by x² + 5.

Quotient will be x² + 2x + 3 and the remainder will be 2x - 8. So, p=2 and q=-8.

For a cubic polynomial ax³ + bx² + cx + d, the sum of products of zeros taken two at a time is c/a.

Here, c/a = -13/2.

A is false because if we let y=x², the polynomial becomes y²+4y+5. The discriminant is 4² - 4(1)(5) = 16-20 = -4 < 0, so there are no real values for y, and thus no real values for x. R is false as the degree is the maximum number of real zeros, not necessarily the exact number.

If x+a is a factor, then x = -a is a zero of the polynomial.

Substitute x = -a into the polynomial: 2(-a)² + 2a(-a) + 5(-a) + 10 = 0 => 2a² - 2a² - 5a + 10 = 0 => -5a + 10 = 0 => 5a = 10 => a = 2.

If -2 and 5 are zeros, then the quadratic polynomial is k(x² - ((-2)+5)x + (-2)(5)) = k(x² - 3x - 10).

Since k can be any non-zero real number, there are infinitely many such polynomials.

alpha + beta = -b/a and alpha * beta = c/a. alpha/beta + beta/alpha = (alpha² + beta²) / (alpha*beta).

alpha² + beta² = (alpha+beta)² - 2alpha*beta = (-b/a)² - 2(c/a) = b²/a² - 2c/a = (b² - 2ac)/a².

So, (b² - 2ac)/a² / (c/a) = (b² - 2ac)/a² * (a/c) = (b² - 2ac) / ac.

If (x-3) is a factor, then x=3 must be a zero of the modified polynomial.

Let P(x) = x² - 5x + 4. P(3) = 3² - 5(3) + 4 = 9 - 15 + 4 = -2. To make it 0, we need to add 2.

For x² + x + 1, alpha + beta = -1 and alpha * beta = 1.

For the new polynomial, sum of zeros = 1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = -1/1 = -1. Product of zeros = (1/alpha)*(1/beta) = 1/(alpha*beta) = 1/1 = 1.

The new polynomial is x² - (sum)x + (product) = x² - (-1)x + 1 = x² + x + 1.