Class 9 Mathematics Chapter 2 Polynomials

Key Concept: Term Count

c) 4

[Solution Description] A polynomial’s terms are separated by addition or subtraction signs. In the given polynomial $3x^4 + 5x^2 – x + 6$, the terms are $3x^4$, $5x^2$, $-x$, and $6$. Therefore, there are 4 separate terms in this polynomial.

Key Concept: Complex Substitution

c) -11

[Solution Description] Substitute $x = -1$ into $p(x)$:

$p(-1) = 3(-1)^3 – 2(-1)^2 + (-1) – 5$

$= 3(-1) – 2(1) – 1 – 5$

$= -3 – 2 – 1 – 5$

$= -11$.

Key Concept: Factor Theorem Application

a) Yes, it is a factor.

[Solution Description]

To determine if $x-3$ is a factor, use the Factor Theorem by evaluating the polynomial at $x=3$:

$p(3) = 3^3 – 6 \cdot 3^2 + 11 \cdot 3 – 6 = 27 – 54 + 33 – 6 = 0$

Since $p(3) = 0$, $x-3$ is indeed a factor of the polynomial.

Key Concept: Zero Calculation

b) 3

[Solution Description] The zero of a linear polynomial $ax + b$ is given by the formula $x = -\frac{b}{a}$. For the polynomial $p(x) = 7x – 21$, we have $a = 7$ and $b = -21$.

Substitute these values into the formula: $x = -\frac{-21}{7} = \frac{21}{7} = 3$

Therefore, the zero of the polynomial is $3$.

Key Concept: Identity Recognition

c) $x^2 – y^2 = (x + y)(x – y)$

[Solution Description]

To solve this question, we need to recognize which algebraic identity can be applied.

The given expression is $49x^2 – 9y^2$.

This fits the form of the difference of squares: $x^2 – y^2 = (x + y)(x – y)$

Here, $x = 7x$ and $y = 3y$, so applying the identity gives: $49x^2 – 9y^2 = (7x + 3y)(7x – 3y)$

Therefore, the applicable identity is $x^2 – y^2 = (x + y)(x – y)$.

Key Concept: Zero of General Linear Polynomial

c) -2

[Solution Description] To find the zero of the linear polynomial $p(x) = 7x + 14$, we use the formula for finding the zero: $x = -\frac{b}{a}$ where $a = 7$ and $b = 14$.

Thus, the calculation becomes: $x = -\frac{14}{7}$

Solving gives: $x = -2$

Hence, the zero of the polynomial is $-2$.

Key Concept: Number of Zeroes

c) 3

[Solution Description] The maximum number of zeroes a polynomial can have is equal to its degree. Since a cubic polynomial has a degree of 3, it can have at most 3 zeroes.

Key Concept: Zero Calculation

b) Zero is -2 and $x + 2$ is a factor

[Solution Description]

A zero of the polynomial $q(x)=2x+4$ satisfies $q(x)=0$.

Solve for $x$: $2x+4=0 $2x=-4$ $x=-2$

Since $q(-2) = 0$, by the Factor Theorem, $x + 2$ is a factor.

Key Concept: Zero Determination

a) 3

[Solution Description] To find the zero of the polynomial $p(x) = 3x – 9$, we set $3x – 9 = 0$. Solving for $x$:

$3x – 9 = 0$

$3x = 9$

$x = \frac{9}{3}$

$x = 3$

So, the zero of the polynomial is $x = 3$.

Key Concept: Identity Verification, Advanced Reasoning

a) True; $(x^3 + y^3) = (x+y)(x^2 – xy + y^2)$

[Solution Description]

We begin with the given expression $x^3 + y^3 – 3xy(x+y)$. Notice the term $-3xy(x+y)$ hints at employing the identity related to $x^3 + y^3 + z^3 – 3xyz$. This is remembered by Identity VIII.

Rewrite the expression:

$x^3 + y^3 – 3xy(x+y)  = x^3 + y^3 + (-z)^3 – 3xy(-z)$,

$\text{where } z = -(x+y)$.

Applying Identity VIII: $(x + y + z)\left(x^2 + y^2 + z^2 – xy – yz – zx\right)$.

Substituting $z = -(x+y)$: $$(x + y – (x+y))\left(x^2 + y^2 + (-(x+y))^2 – xy – y(-(x+y)) – x(-(x+y))\right),$$

Simplifying further leads to $0$. Similarly:

$$(x+y)\left(x^2 – xy + y^2\right).$$

Therefore, the equivalent form matches the product form for $x^3 + y^3$

The verified conversion supports the transformation being true.

Key Concept: Complex Polynomial Analysis, Real-World Application

a) $2h – 1$

[Solution Description] Let the height be $h$, the width be $w = 2h – 1$, and the depth be $h$. The volume, therefore, is $V = w \cdot h \cdot d = (2h – 1) \cdot h \cdot h$.

Simplifying, $V = (2h – 1)h^2 = 2h^3 – h^2$

We see this matches the provided volume. Thus the expression for width $w$ is $2h – 1$.

Key Concept: Zero Calculation

c) 4

[Solution Description]

To find the zero of a linear polynomial $p(x) = ax + b$, we set it equal to zero and solve for $x$.

For $p(x) = 3x – 12$, set the expression to zero: $3x – 12 = 0$

Adding 12 to both sides, we get: $3x = 12$

Dividing both sides by 3, we obtain: $x = \frac{12}{3} = 4$

Therefore, the zero of the polynomial is $x = 4$.

Key Concept: Identify Degree

d) 4

[Solution Description] The degree of a polynomial is defined as the highest power of the variable in the polynomial. In this case, the highest power of $x$ is 4, which makes the degree of the polynomial 4.

Key Concept: Zero Existence

b) No

[Solution Description]

A quadratic polynomial $ax^2 + bx + c$ will have real zeros if its discriminant $D = b^2 – 4ac$ is non-negative. For $p(x) = x^2 + 4x + 5$, we calculate:

$D = 4^2 – 4 \cdot 1 \cdot 5 = 16 – 20 = -4$

Since the discriminant is negative ($-4 < 0$), the polynomial has no real zeroes.

Key Concept: Basic Identities

b) $x^2 – 2xy + y^2$

[Solution Description] The algebraic identity for $(x – y)^2$ is given by the formula:

$(x – y)^2 = x^2 – 2xy + y^2$

So, the correct expansion is $x^2 – 2xy + y^2$.

Key Concept: Identify Monomial

a) $5x^3$

[Solution Description] A monomial is a polynomial with only one term. Among the given options, we need to identify which expression has only one term.

Key Concept: Degree of Polynomial

c) 3

[Solution Description] The degree of a polynomial is determined by the highest power of the variable. In this case, the highest power of $x$ is 3 in the term $5x^3$. Therefore, the degree of the polynomial is 3.

Key Concept: Multiple Zero Verification

d) Both are zeroes

[Solution Description] To verify, we substitute each number into $q(x)$.

– For $x = -1$: $q(-1) = (-1)^2 – (-1) – 2 = 1 + 1 – 2 = 0$.

– For $x = 2$: $q(2) = 2^2 – 2 – 2 = 4 – 2 – 2 = 0$.

Both evaluations result in 0, so both $-1$ and $2$ are zeroes of the polynomial.

Key Concept: Zero Calculation

c) 3

[Solution Description] A zero of the polynomial $p(x) = x – 3$ is a value of $x$ such that $p(x) = 0$. Setting $x – 3 = 0$, we solve for $x$: $x = 3$

Key Concept: Factor Identification

c) $x – 1$

[Solution Description] To determine if $(x – 1)$ is a factor, substitute $x = 1$ into the polynomial $p(x)$.

First, calculate $p(1) = 1^3 – 6 \cdot 1^2 + 11 \cdot 1 – 6 = 1 – 6 + 11 – 6 = 0$. Since $p(1) = 0$, by the Factor Theorem, $(x – 1)$ is a factor.

Key Concept: Identify Trinomial

b) $2x^2 + 3x + 5$

[Solution Description] A trinomial has three terms. Option b) $2x^2 + 3x + 5$ consists of exactly three terms: $2x^2$, $3x$, and $5$. Hence, it is a trinomial.

Key Concept: Factor Pair Identification

a) 1 and 6

[Solution Description] For the polynomial $3x^2 + 7x + 2$, we need to find two numbers that sum to 7 and multiply to the product of the leading coefficient and constant term, which is $3 \cdot 2 = 6$. The correct pair of numbers is 1 and 6 because $1 + 6 = 7$ and $1 \cdot 6 = 6$.

Key Concept: Identity Derivation, Real-World Application

b) $2197$

[Solution Description] Start by recognizing that we can express $x^2 + y^2$ as follows: $x^2 + y^2 = (x + y)^2 – 2xy$

Substituting the known values: $x^2 + y^2 = 5^2 – 2 \cdot 6 = 25 – 12 = 13$

Now, calculate $(x^2 + y^2)^3$: $(13)^3 = 2197$

Therefore, the derived simplified form is $2197$.

Key Concept: Identify Degree

c) 4

[Solution Description] The degree of a polynomial is the highest power of the variable in the polynomial. For the polynomial $3x^4 + 2x^3 – x + 7$, the highest power of $x$ is 4.

Key Concept: Polynomial in Multiple Variables, Polynomial Term Limit

d) Its highest degree term is 4.

[Solution Description]

A polynomial can have multiple variables, and each term’s degree is the sum of the exponents of its variables. In $g(x, y)$, the term with the highest total degree is $xy^3$, where the degree is $1 + 3 = 4$. Similarly, terms like $x^3y$ also have a total degree of $3 + 1 = 4$. For a single polynomial expression, the maximum number of terms depends on how it is constructed but doesn’t exceed the combinations of powers in multivariable cases unless restricted by the form.

Key Concept: Factor Identification

b) $x^2 + 4x + 4$

[Solution Description] To determine whether $x + 2$ is a factor of a polynomial, we apply the Factor Theorem: If $p(-2) = 0$, then $x + 2$ is a factor. For the polynomials given:

– For $p(x) = x^2 + 4x + 4$:

$p(-2) = (-2)^2 + 4(-2) + 4$

$= 4 – 8 + 4$

$= 0$

Since $p(-2) = 0$, $x + 2$ is a factor.

– Other options do not satisfy this condition.

Key Concept: Degree in Real-world Context, Degree of Polynomial Function

c) 4

[Solution Description] To determine the degree of this polynomial function, identify the highest power of $t$ in the expression:

$V(t) = 10t^4 – 6t^3 + 4t^2 – 3t + 8$

Here, the term with the highest degree is $10t^4$, which indicates that the degree of the polynomial is $4$.

Key Concept: Quadratic Zeroes

a) 1 and 2

[Solution Description] To find the zeroes of the polynomial, we need to solve $x^2 – 3x + 2 = 0$. We can factor this as $(x – 1)(x – 2) = 0$. Setting each factor equal to zero gives $x = 1$ and $x = 2$. Therefore, the zeroes of the polynomial are 1 and 2.

Key Concept: Evaluate Complex Products, Advanced Reasoning

d) $4$

[Solution Description] Let’s denote $a = 1000$, so we have:

$(1001)^3 = (a + 1)^3$

$(999)^3 = (a – 1)^3$

We need to evaluate $(a + 1)^3 – (a – 1)^3$. Using Identity VIII:

$(a + 1)^3 – (a – 1)^3 = ((a + 1) – (a – 1))(a^2 + 1^2 + (-1)^2 – (a + 1)(a – 1))$

Simplifying gives:

$= (2)(a^2 + 1 + 1 – (a^2 – 1))$

$= 2(2)$

$= 4$

Hence, the value of $(1001)^3 – (999)^3$ is $4$.

Key Concept: Polynomial Formulation, Polynomial Degree in Real World

c) 4

[Solution Description]

The polynomial provided is $P(x) = -x^4 + 5x^3 – 10x + 25$. The degree corresponds to the highest power of $x$ present in the expression. In this polynomial, the term $-x^4$ holds the highest exponent of 4. Therefore, the degree of the polynomial is 4. The degree signifies the type of growth or decay characteristic in the company’s profit related to the investments.

Key Concept: Identity Application

c) 9801

[Solution Description]

We can use Identity II: $(x – y)^2 = x^2 – 2xy + y^2$. Let $x = 100$ and $y = 1$.

Step 1: Substitute the values into the identity.

$(99)^2 = (100 – 1)^2 = 100^2 – 2(100)(1) + 1^2$

Step 2: Calculate each term.

$100^2 = 10000, \quad 2(100)(1) = 200, \quad 1^2 = 1$

Step 3: Combine the terms.

$10000 – 200 + 1 = 9801$

The value of $(99)^2$ is 9801.

Key Concept: Zero Identification

b) 1 and 3

[Solution Description] To find the zeros of the polynomial $p(x) = x^2 – 4x + 3$, we solve the equation $x^2 – 4x + 3 = 0$. We can factorize it as $(x-1)(x-3) = 0$. Setting each factor to zero gives $x-1=0$ or $x-3=0$ leading to $x=1$ or $x=3$. Hence, the zeros are $x=1$ and $x=3$.

Key Concept: Middle Term Splitting

c) $(3x – 2)(2x + 5)$

[Solution Description] To factorize $6x^2 + 11x – 10$, we need to split the middle term. We look for two numbers whose product is $6 \times (-10) = -60$ and sum is $11$. These numbers are $15$ and $-4$.

Rewrite the expression as $6x^2 + 15x – 4x – 10$.

Group terms: $(6x^2 + 15x) + (-4x – 10)$.

Factor each group: $3x(2x + 5) – 2(2x + 5)$.

Since both groups contain a common factor of $(2x + 5)$, the expression can be written as: $(3x – 2)(2x + 5)$.

Key Concept: Polynomial Classification

a) $3x^2 + 4$

[Solution Description] A binomial is a polynomial with exactly two terms. Therefore, we need to identify the expression that has only two distinct terms.

Key Concept: Linear Form

c) $3x + 1$

[Solution Description] A linear polynomial in one variable is of the form $ax + b$, where $a$ and $b$ are constants.

Key Concept: Identity Application

a) $x^2 + 10x + 25$

[Solution Description] To expand $(x + 5)^2$, we use Identity I: $(x + y)^2 = x^2 + 2xy + y^2$.

Here, $y = 5$. So: $(x + 5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$

Key Concept: Identity Application

b) $x^2 – y^2 = (x + y)(x – y)$

[Solution Description] The expression $x^2 – 9$ is a difference of squares, which matches Identity III: $x^2 – y^2 = (x + y)(x – y)$. Here, $y = 3$, so $x^2 – 9 = (x + 3)(x – 3)$.

Key Concept: Zero Verification

a) Yes, $x = 2$ is a zero

[Solution Description] A number $\alpha$ is a zero of a polynomial $q(x)$ if $q(\alpha) = 0$. Substitute $x = 2$ into the polynomial:

$q(2) = 2^2 – 3(2) + 2$

$= 4 – 6 + 2$

$= 0$

Since $q(2) = 0$, $x = 2$ is a zero of the polynomial $q(x)$.

Key Concept: Zero of Quadratic Polynomial

a) 2 and 3

[Solution Description] To find the zeroes of the polynomial $q(x) = x^2 – 5x + 6$, we solve the equation $q(x) = 0$:

Factorize the polynomial: $x^2 – 5x + 6 = (x – 2)(x – 3)$

Setting each factor to zero gives the solutions:

$x – 2 = 0 \quad \Rightarrow \quad x = 2$

$x – 3 = 0 \quad \Rightarrow \quad x = 3$

Therefore, the zeroes of the polynomial are 2 and 3.

Key Concept: Identity Application

c) 11025

[Solution Description] To evaluate $105^2$, we can use Identity I: $(x + y)^2 = x^2 + 2xy + y^2$. In this case, let $x = 100$ and $y = 5$.

$105^2 = (100 + 5)^2 = 100^2 + 2 \cdot 100 \cdot 5 + 5^2$

$= 10000 + 1000 + 25$

$= 11025$

Key Concept: Non-zero Constant Polynomial

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

The assertion states that the constant polynomial 9 has no zero. A non-zero constant polynomial does not equal zero for any real number $x$, making this statement true. The reason supports this by stating that substituting any value for $x$ into the polynomial 9 results in the value 9, which is indeed not zero. Therefore, both the assertion and the reason are true, and the reason correctly explains why the assertion is true.

Key Concept: Zero Calculation

d) $-3$

[Solution Description] To find the zero of $p(x)$, we set $p(x) = 0$. Thus, $x + 3 = 0$, which gives $x = -3$. Since substituting $x = -3$ into $p(x)$ gives $0$, $x + 3$ is indeed a factor.

Key Concept: Degree Identification

b) 5

[Solution Description] The degree of a polynomial is determined by the highest power of the variable in the polynomial. In this case, the term with the highest power of $x$ is $4x^5$, so the degree of the polynomial is 5.

Key Concept: Perimeter Formula

b) 28 units

[Solution Description]

To find the perimeter of a square, we use the formula $4x$, where $x$ is the side length of the square.

Substituting $x = 7$, we get: $\text{Perimeter} = 4 \times 7 = 28 \text{ units}$.

Thus, the perimeter of the square is 28 units.

Key Concept: Identity Recognition

c) Identity VI: $(x+y)^3$

[Solution Description] To expand $(a + b)^3$, we use Identity VI: $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$.

Thus, it can be expanded as: $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$