Class 10 Mathematics Chapter 2 Polynomials

Key Concept: Coefficient Comparison

(c) 3/2

Solution Description:

To determine the product of the zeroes of the given polynomial, we use the relationship between the coefficients and the zeroes. The product of the zeroes of a quadratic polynomial is obtained directly from the given expression. By identifying the required terms, we apply the standard formula for finding this value. Substituting the correct coefficients, we simplify the expression carefully to reach the final result. After performing the necessary calculations, we conclude that the product of the zeroes for this polynomial is as given in the option. Thus, the correct answer is obtained following the appropriate mathematical approach.

Key Concept: Degree Identification

c) 4

Solution Description:
The degree of a polynomial is the highest power of x that appears in the polynomial. Here, the highest power of x is 4.

Key Concept: Parabola Orientation

(b) Opens upwards

Solution Description:

The given equation represents a quadratic function in the standard form. In a quadratic equation, the direction of the graph depends on the coefficient of the squared term. If the coefficient is positive, the graph opens upwards; if negative, it opens downwards. Here, the coefficient of the squared term is positive, meaning the graph forms a U-shaped curve. This indicates that the function has a minimum point and extends infinitely in the upward direction. Since the leading coefficient determines the orientation, we conclude that the graph of the given function opens upwards.

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Key Concept: Product of Zeroes

(c) 4

Solution Description:

The product of the zeroes of a cubic polynomial can be determined using a relationship between the coefficients and the roots. In this case, the polynomial is cubic, and by applying Vieta’s formulas, we can find that the product of the zeroes is linked to the constant term and the leading coefficient. The formula for the product of the zeroes gives a direct connection to these values, helping us identify the result without explicitly solving for the roots. Through this approach, we can efficiently compute the product of the zeroes, which corresponds to a given value in the context of the polynomial.

Key Concept: Zeroes Identification

(a) 3, 4

Solution Description:

Key Concept: Degree Identification

c) 4

Solution Description:
The degree of a polynomial is the highest power of the variable in the expression. Here, the highest power of x is 4.

Key Concept: Sum of Zeroes

(b) $\frac{5}{2}$

Solution Description:

To find the sum of the zeroes of the quadratic polynomial, we use the standard property that states the sum of the roots of a quadratic equation is given by the ratio of the coefficient of the middle term to the coefficient of the leading term. Observing the given polynomial, we identify the required coefficients. Dividing the middle term’s coefficient by the leading term’s coefficient gives the result. Simplifying the fraction, we get the final value as five divided by two. Hence, the sum of the zeroes of the given quadratic polynomial is five over two.

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Key Concept: Polynomial Construction

a) $x^2− 5x − 6$

Solution Description:
To construct a quadratic polynomial given the sum and product of its roots, use the general form:

$x^2$− (sum of roots)x + product of roots. Here, the sum of roots is 5 and the product is -6. Thus, the polynomial is

$x^2 − 5x − 6$

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Key Concept: Zeroes and Coefficients

c) 4

Solution Description:
The sum of the zeroes $\alpha + \beta = -\frac{b}{a}$ ​. Given zeroes are 4 and 13 $\frac{1}{3}$ So,

$4 + \frac{1}{3} = -\frac{-7}{3} = \frac{7}{3}$
Therefore, converting 4 to a fraction: This confirms that the zeroes satisfy this condition. Now for the product of the zeroes

$\alpha \beta = \frac{k}{3}$​, so:

$4 \cdot \frac{1}{3} = \frac{k}{3}$

​Hence, solving for k:

$\frac{k}{3} = \frac{4}{3}$

Multiply both sides by 3: k = 4. Thus, option ‘c’ is correct.

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Key concept: Rational Root Theorem, Sum and Product of Zeroes

(a) Zeroes at 1, 2, 3 with coefficient sum zero

Solution Description:

The given cubic polynomial has zeroes at 1, 2, and 3, which means these values satisfy the equation.  The product of the zeroes also aligns with the constant term. Additionally, the sum of all coefficients of the polynomial results in zero, indicating a specific symmetry in its structure. This relationship between the zeroes and coefficients follows fundamental properties of polynomials, confirming that the polynomial can be factored accordingly and fully analyzed through algebraic principles.

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Key Concept: Zero Verification

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Key Concept: Zero Sum Product Analysis, Advanced Reasoning

(c) 11

Solution Description:

We are given that α, β, and γ are the zeroes of the cubic polynomial $p(x) =x^3− 6x^2+ 11x − 6,$ We are tasked with calculating the expression αβ+βγ+γα.

To solve this, we use Vieta’s formulas, which relate the coefficients of a polynomial to sums and products of its roots. In this case, Vieta’s relations provide us with the sum and product of the roots. By identifying the required sum of products, we can directly compute the desired value.

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Key Concept: Zeroes and Coefficients

(a) $\frac{2}{3}$

Solution Description:

For a cubic polynomial, the sum of the products of its zeroes taken two at a time is derived from its coefficients. Given the polynomial, its structure determines this sum based on a standard formula. By identifying the appropriate coefficient relationships, we extract the required value. Applying the formula, we find that the sum of the products of the zeroes taken two at a time is \[
\frac{2}{3}
\]. This result is obtained by carefully analyzing the polynomial’s terms and using known mathematical properties. Hence, the answer is derived directly from the given polynomial’s coefficient arrangement.

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Key Concept: Zeroes of Polynomial

c) 4

Solution Description:
According to the properties of polynomials, a polynomial of degree n can have at most n zeroes. Therefore, a polynomial of degree 4 can have at most 4 zeroes.

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Key Concept: Polynomial Evaluation

b) 0

Solution Description:
Substitute x = 3 into the polynomial and evaluate: $$p(3) = (3)^2 − 3 − 6 = 9 − 3 − 6 = 0$$ Thus, the value of the polynomial at x = 3 is 0.

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Key Concept: Zero Calculation in Context, Zero Relationship

d) 0

Solution Description:

$p(x) = x(2x^2 – 9x + 12)$

Next, factor the quadratic: Using the quadratic formula:

$ax^2 + bx + c = 0 \quad \Rightarrow \quad x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

$x = \frac{-9 \pm \sqrt{81 – 96}}{4}$

Simplifying the square root: $x = 9 \pm \sqrt{-154}$

Since there are no real roots from the quadratic part, consider x = 0

The only real zero is at. x = 0 Hence, the sum of all zeroes, considering complex zeroes cancel each other out and only real ones remain, is 0.

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Key Concept: Zero Identification

b) 2, 3

Solution Description:
To find the zeroes, solve the equation $x^2−5x+ 6 = 0$. Factorize it: $x^2−5x+ 6 =(x − 2)(x − 3)$ Set each factor to zero: x − 2 = 0 ⇒ x = 2 x − 3 = 0 ⇒ x = 3 Thus, the zeroes are x = 2
and x = 3.

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Key Concept: Graph Analysis

a) -4

Solution Description:
The zero of the polynomial occurs where y = 0. Set 2x + 8 = 0:
2x + 8 = 0

2x = −8
x =−8/2
= −4
Hence, the zero of the polynomial is -4.

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Key Concept: Polynomial Evaluation

(a) 0

Solution Description:

Substitute x = 1 into the polynomial $p(x) =x^2 + 2x − 3:p(1) = 1^2 + 2(1) − 3= 1 + 2 − 3= 0$
So, the value of the polynomial at x = 1 is 0.

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Key Concept: Polynomial Form

b) $ax^3+bx^2+cx+d$

Solution Description:

A cubic polynomial is generally expressed in the form $ax^3+bx^2+cx+d$. This includes all terms up to degree three with coefficients.

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