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Unit Test (Solutions): Polynomials | Mathematics (Maths) Class 9 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

Question numbers 1 to 5 carry 1 mark each.

Question numbers 6 to 8 carry 2 marks each.

Question numbers 9 to 11 carry 3 marks each.

Question number 12 & 13 carry 5 marks each

Q1: Which of the following is a polynomial in one variable? (1 Mark)

(i) $x^2 + \frac{1}{x}$

(ii) $2 x 2 + 3 x + 5$

(iii) $y 2 + x y$

(iv) 2/x + x

Ans: (ii)
A polynomial in one variable should not have any variable terms in the denominator or fractional exponents. It must have powers of a single variable with non-negative integers as exponents.

Option (i) is not a polynomial because $\frac{1}{x}$ includes a negative exponent.

Option (ii) is a polynomial in one variable $x$ , and it satisfies all conditions of a polynomial.

Option (iii) has two variables ( $y$ and $x$ ), so it’s not a polynomial in one variable.

Option (iv) contains $\frac{2}{x}$ , which has a negative exponent.

Q2: What is the degree of the polynomial $5 x 3 + 4 x 2 - x + 2$ ? (1 Mark)
(i) 1
(ii) 2
(iii) 3
(iv) 4

Ans: (iii)
The degree of a polynomial is the highest power (exponent) of the variable in the polynomial. In the polynomial $5 x 3 + 4 x 2 - x + 2$ 5x³+4x²−x+2, the highest power of $x$ is 3.
Thus, the degree of this polynomial is 3.

Q3: Which of the following is a zero of the polynomial $p (x) = x - 3$ ? (1 Mark)
(i) 1
(ii) 0
(iii) 3
(iv) -3

Ans: (iii)
A zero of a polynomial is a value of $x$ x that makes the polynomial equal to zero. To find the zero of $p (x) = x - 3$ , we set the polynomial equal to 0:

$x - 3 = 0$ $x = 3$
x=3

Thus, 3 is the zero of the polynomial.

Q4: What is the degree of the constant polynomial $7$ 7? (1 Mark)
(i) 0
(ii) 1
(iii) Not defined
(iv) 7

Ans: (i)
A constant polynomial is a polynomial with no variable terms (it is just a number). The degree of a constant polynomial is always 0 because there are no variables with exponents.
Thus, the degree of the constant polynomial $7$ 7 is 0.

Q5: The factor theorem states that $x - a$ x−a is a factor of the polynomial $p(x)$ p(x) if: (1 Mark)
(i) $p(a) = 1$ p(a)=1
(ii) $p(a) = a$ p(a)=a
(iii) $p(a) = 0$ p(a)=0
(iv) $p(a) \neq 0$

Ans: (c)
According to the factor theorem, $x - a$ is a factor of a polynomial $p (x)$ if and only if $p (a) = 0$ . This means that if you substitute $a$ into the polynomial and the result is zero, then $x - a$ is a factor of the polynomial.

Q6: Find the value of $k$ , if $x + 2$ is a factor of $p (x) = 3 x 3 + 2 x 2 - 5 x + k$ . (2 Marks)

Ans: Since $x + 2$ is a factor of $p (x)$ , by the Factor Theorem, $p (- 2) = 0$ .

Now, substitute $x = - 2$ into the polynomial $p (x) = 3 x 3 + 2 x 2 - 5 x + k$ :

Unit Test (Solutions): Polynomials | Mathematics (Maths) Class 9

Since $p (- 2) = 0$ , we can set up the equation:

$- 6 + k = 0$

Solve for $k$ :

$k = 6$

Thus, the value of $k$ is 6.

Q7: Factorize the quadratic expression $4 x 2 + 7 x - 2.$ $(2 Marks)$ $4x^2 + 7x - 2$

Ans: To factorize the quadratic expression $4 x 2 + 7 x - 2$ , we need to split the middle term (7x) into two terms whose product is equal to $4 \times - 2 = - 8$ , and whose sum is equal to 7.

The two numbers that satisfy these conditions are 8 and -1 (since $8 \times -1 = -8$ and $8 + (- 1) = 7$ ).

So, we rewrite the expression as:

$4 x 2 + 8 x - x - 2$

Now, group the terms:

$(4 x 2 + 8 x) - (x + 2)$

Factor out the common terms:

$4 x (x + 2) - 1 (x + 2)$

Now, factor out $(x + 2)$ :

$(x + 2) (4 x - 1)$

Thus, the factorized form of $4 x 2 + 7 x - 2$ is:

$(x + 2) (4 x - 1)$

Q8: Evaluate $108 \times 109$ without multiplying directly.

Ans:

To evaluate $108 \times 109$ without direct multiplication, we can use the distributive property.

Rewrite the numbers:

108 \times 109 = (100 + 8) \times (100 + 9)

Apply the distributive property :

= 100^2 + (8 + 9) \times 100 + (8 \times 9)

Calculate each term:

$1002 = 10000$

$8 + 9 = 17$ so $17 \times 100 = 1700$

$8 \times 9 = 72$

Combine the results:

10000 + 1700 + 72 = 11772

Q9: Use the Factor Theorem to determine whether $g (x) = x - 2$ is a factor of $p (x) = 2 x 3 + x 2 - 2 x - 1$ . (3 Marks)

Ans:
To check if $g (x) = x - 2$ is a factor of $p (x) = 2 x 3 + x 2 - 2 x - 1$ using the Factor Theorem, we first evaluate $g (2)$ .

Evaluate $g (2)$ :

g (2) = 2 - 2 = 0

Since $g (2) = 0$ , $x - 2$ is a candidate for being a factor.

Perform polynomial long division of $p (x)$ by $g (x) = x - 2$ :

Divide the leading term $2 x 3$ by $x$ to get $2 x 2$ .
Multiply $2 x 2$ by $g (x)$ : $2 x 2 (x - 2) = 2 x 3 - 4 x 2$
Subtract this from $p (x)$ : $(2 x 3 + x 2 - 2 x - 1) - (2 x 3 - 4 x 2) = 5 x 2 - 2 x - 1$
Now, divide the leading term $5 x 2$ by $x$ to get $5 x$ .
Multiply $5 x$ by $g (x)$ : $5 x (x - 2) = 5 x 2 - 10 x$
Subtract: $(5 x 2 - 2 x - 1) - (5 x 2 - 10 x) = 8 x - 1$
Finally, divide $8 x 8$ by $x$ to get $8$ .
Multiply $8$ by $g (x)$ : $8 (x - 2) = 8 x - 16$
Subtract: $(8 x - 1) - (8 x - 16) = 15$

Since the remainder is $15$ 15, which is not $0$ 0, we conclude that $g (x) = x - 2$ is not a factor of $p (x)$ .
Q10: Expand $(5 x + 3 y - 4 z) 2$ .(3 Marks)
Ans:

Identify the terms:

(5 x + 3 y - 4 z) 2 = [5 x + 3 y + (- 4 z)] 2

Apply the expansion formula:

= (5 x) 2 + (3 y) 2 + (- 4 z) 2 + 2 (5 x) (3 y) + 2 (3 y) (- 4 z) + 2 (- 4 z) (5 x)

Calculate each term:

$(5 x) 2 = 25 x 2$

$(3 y) 2 = 9 y 2$

$(- 4 z) 2 = 16 z 2$

$2 (5 x) (3 y) = 30 x y$

$2 (3 y) (- 4 z) = - 24 y z$

$2 (- 4 z) (5 x) = - 40 z x$

Combine the results:

25 x 2 + 9 y 2 + 16 z 2 + 30 x y - 24 y z - 40 z x

Q11: If $a + b + c = 0$ , show that $a 3 + b 3 + c 3 = 3 a b c$ (3 Marks)
$Ans:$

Given the condition $a + b + c = 0$ , we can use the identity for the sum of cubes:

a 3 + b 3 + c 3 - 3 a b c = (a + b + c) (a 2 + b 2 + c 2 - a b - a c - b c)

Since $a + b + c = 0$ , we can substitute this into the identity:

a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - ac - bc)

This simplifies to:

a 3 + b 3 + c 3 - 3 a b c = 0

Thus, we can conclude that:

a 3 + b 3 + c 3 = 3 a b c

Q12: Calculate the perimeter of a rectangle whose area is 25x² – 35x + 12. (5 Marks)

Ans:

Given,

Area of rectangle = 25x² – 35x + 12

We know, area of rectangle = length × breadth

So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.

25x² – 35x + 12 = 25x²– 15x – 20x + 12

=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x² – 35x + 12 = (5x – 3)(5x – 4)

So, the length and breadth are (5x – 3)(5x – 4).

Now, perimeter = 2(length + breadth)

So, perimeter of the rectangle = 2[(5x – 3)+(5x – 4)]

= 2(5x – 3 + 5x – 4) = 2(10x – 7) = 20x – 14

So, the perimeter = 20x – 14

Q13: Factorize $4 x 2 + 2 y 2 + z 2 - 4 x y - 4 y z + 4 x z$ (5 Marks)
Ans:

To factor the expression $4 x 2 + 2 y 2 + z 2 - 4 x y - 4 y z + 4 x z$ , we can rearrange and group the terms.

Rearranging the expression:

4 x 2 - 4 x y + 4 x z + 2 y 2 - 4 y z + z 2

Group terms:

= (4 x 2 - 4 x y + 4 x z) + (2 y 2 - 4 y z + z 2)

Factor out common terms from each group:

From $4 x 2 - 4 x y + 4 x z$ , factor out 4: $= 4 (x 2 - x y + x z)$

From $2 y 2 - 4 y z + z 2$ , we can rearrange and factor:

Unit Test (Solutions): Polynomials | Mathematics (Maths) Class 9
Putting it together: =4(x²−xy+xz)+2(y²−2yz+z²)

Factoring further:

Notice $x 2 - x y + x z$ can be recognized as: $= (x - y) (x + z)$

And for $y 2 - 2 y z + z 2$ : $= (y - z) 2$ ^{= (y - z)^2}

Finally, we can express the entire factorization as:

= (2 (x - y + z)) (x + (y - z))

The document Unit Test (Solutions): Polynomials | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.

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FAQs on Unit Test (Solutions): Polynomials - Mathematics (Maths) Class 9

1. What is a polynomial and how is it defined?

Ans. A polynomial is a mathematical expression that consists of variables raised to non-negative integer powers and coefficients. It is typically written in the form $ a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 $, where $ a_n, a_{n-1}, ..., a_0 $ are constants (coefficients), $ x $ is the variable, and $ n $ is a non-negative integer that represents the degree of the polynomial.

2. How do you add and subtract polynomials?

Ans. To add or subtract polynomials, combine like terms. Like terms are terms that have the same variable raised to the same power. For example, to add $ (3x^2 + 4x + 5) + (2x^2 + 3) $, you add the coefficients of like terms: $ (3x^2 + 2x^2) + 4x + (5 + 3) = 5x^2 + 4x + 8 $.

3. What is the difference between a monomial, binomial, and trinomial?

Ans. A monomial is a polynomial with just one term, such as $ 3x^2 $. A binomial has two terms, like $ 4x + 5 $. A trinomial consists of three terms, for example, $ x^2 + 2x + 1 $. The number of terms distinguishes these types of polynomials.

4. How can polynomials be factored?

Ans. Polynomials can be factored by finding their greatest common factor (GCF) or by using methods such as grouping, the difference of squares, or special factoring formulas (like $ a^2 - b^2 = (a - b)(a + b) $). For example, the polynomial $ x^2 - 9 $ can be factored as $ (x - 3)(x + 3) $.

5. What are the methods to solve polynomial equations?

Ans. Polynomial equations can be solved using various methods, such as factoring, using the quadratic formula (for second-degree polynomials), synthetic division, or numerical methods like the Newton-Raphson method for higher-degree polynomials. Each method is chosen based on the degree and complexity of the polynomial.

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