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CBSE Questions for Class 9 Maths Polynomials Quiz 2 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 2
Use the identity $$ (a+b)(a-b) = a^2-b^2$$ to evaluate:
$$21\times 19 $$.
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0%
$$389$$
0%
$$399$$
0%
$$289$$
0%
$$429$$
Explanation
We know, $$ 21 \times 19 = (20+ 1) \times (20 - 1) $$ .
Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$,
where $$ a = 20 , b = 1 $$,
we get,
$$ 21 \times 19 = (20+ 1) \times (20 - 1) = { 20 }^{ 2 }-{ 1 }^{ 2 } = 400 - 1 = 399 $$ .
Therefore, option $$B$$ is correct.
What is the degree of the following polynomial expression:
$$\dfrac{4}{3}x^{7} - 3x^{5} + 2x^{3} + 1$$
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0%
7
0%
4
0%
5
0%
3
Explanation
Clearly,
the degree of the polynomial expression
$$\dfrac{4}{3}x^{7} - 3x^{5} + 2x^{3} + 1$$ is $$7$$.
The
degree of a polynomial
is the highest
degree
of its monomials (individual terms) with non-zero coefficients. The
degree
of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Find the degree of the following polynomial
$$x^9-x^4+x^{12}+x-2$$
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0%
12
0%
9
0%
4
0%
2
Explanation
The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,
$$x^9-x^4+x^{12}+x-2$$
Since there are 4 terms $$x^9-x^4+x^{12}+x-2$$
, this is a
poly
nomial
and has the highest degree $$x^{12}$$
of all the terms.
Therefore,
Degree is $$12$$
Find the degree of the following polynomial
$$3y+4y^2$$
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0%
y
0%
0
0%
1
0%
2
Explanation
We know that, the degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given polynomial is $$3y+4y^2$$
It is a polynomial in $$y$$ having highest exponent value as $$2$$.
Hence, the degree of the given polynomial is $$2$$.
What is the degree of the following polynomial expression:
$$\dfrac{5}{3} x^{3} + 7x + 16$$
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0%
2
0%
3
0%
1
0%
$$\dfrac{5}{3}$$
Explanation
Clearly,
the degree of the polynomial expression
$$\frac{5}{3} x^{3} + 7x + 16$$ is $$3$$.
7
.
The
degree of a polynomial
is the highest
degree
of its monomials (individual terms) with non-zero coefficients. The
degree
of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Find the degree of the following polynomial:
$$-\dfrac{3}{2}$$
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0%
$$1$$
0%
$$2$$
0%
$$0$$
0%
Cannot be determined
Explanation
Given polynomial can be rewritten as $$-\dfrac32 x^0$$ so, the highest degree of the variable present in the polynomial is $$0$$ hence the degree of the polynomial is equal to $$0.$$
Find the degree of following polynomial
$$4x-\sqrt{5}$$
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0%
$$\dfrac{1}{2}$$
0%
1
0%
2
0%
0
Explanation
The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,
$$4x-\sqrt{5}$$
There are 2 terms $$4p-\sqrt{5}$$
,so it is a
binomial
and the polynomial has variable $$x$$
as the highest degree of all the terms
.
Therefore,
Degree is $$1$$
Find the degree of following polynomial $$5$$
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0%
$$5$$
0%
$$1$$
0%
$$0$$
0%
Constant
Explanation
The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.
Given,
$$5$$
Since there is $$1$$ term $$5$$, it is a
monomial
and the polynomial has no variable.
Therefore,
Degree is $$0$$
State true or false:
A binomial may have degree 5.
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0%
True
0%
False
Explanation
A binomial is an algebraic expression with two terms in addition or subtraction. There is no other condition. So, it is true that a binomial may have degree 5.
$$e.g.\space p(x)=x^{5}-7$$, is a binomial.
Therefore, option A is correct.
Use identities to solve: $$(97)^{2}$$
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0%
$$9,659$$
0%
$$9,409$$
0%
$$9,009$$
0%
$$9,209$$
Explanation
$$(97)^2$$
$$=(100-3)^2$$
using, $$(a-b)^2=a^2-2ab+b^2$$
$$=(100)^2-2(100)3+(3)^2$$
$$=10000-600+9$$
$$=9409$$
Find the zero of the polynomial given below:
$$p(x) = 9x - 3$$.
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0%
$$\dfrac{6}{7}$$
0%
$$\dfrac{1}{2}$$
0%
$$\dfrac{1}{3}$$
0%
$$\dfrac{7}{3}$$
Explanation
Given, $$p(x) = 9x - 3$$.
to find the zero of a polynomial we need to equal that polynomial to zero.
thus,
$$p(x)=0$$
$$\Rightarrow9x-3=0$$
$$\Rightarrow x=\dfrac{3}{9}$$
$$\Rightarrow x=\dfrac{1}{3}$$.
Therefore, option $$C$$ is correct.
If p(x)= $$x^3-4x^2+5x-2$$, then p(2) is:
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0%
$$0$$
0%
$$3$$
0%
$$5$$
0%
$$8$$
Explanation
Let $$f(x)=x^3-4x^2+5x-2$$
Now, $$f(2)=2^3-4\times2^2+5\times2-2$$
$$=8-16+10-2=0$$
Since, $$f(2)=0$$.
Then we can say that, $$2$$ is a zero of $$f(x)=x^3-4x^2+5x-2$$.
Therefore, option $$A$$ is correct.
Solve for $$x$$:
$$(502)^{2}$$
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0%
$$2,52,004$$
0%
$$2,62,104$$
0%
$$2,22,004$$
0%
$$2,52,864$$
Explanation
$$(502)^2$$
$$=(500+2)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(500)^2+2(500)(2)+(2)^2$$
$$=250000+2000+4$$
$$=252004$$
Degree of the polynomial $$4x^4+0x^3+0x^5+5x+7$$ is
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0%
$$4$$
0%
$$5$$
0%
$$3$$
0%
$$7$$
Explanation
Clearly, the highest power of $$x$$ is $$4$$.
Hence, the degree of given polynomial is $$4$$.
Option A is correct.
Use identities to evaluate:
$$(101)^{2}$$
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0%
$$11,601$$
0%
$$12,761$$
0%
$$10,111$$
0%
$$10,201$$
Explanation
$$(101)^2$$
$$=(100+1)^2$$
using, $$(a+b)^2=a^2+2ab+b^2$$
$$=(100)^2+2(100)(1)+1^2$$
$$=10000+200+1$$
$$=10201$$
Evaluate using expansion of $$(a+b)^2$$ or $$(a-b)^2$$ :
$$(9.4)^2$$
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0%
88.36
0%
88.46
0%
89.16
0%
89.56
Explanation
$${ (a+b) }^{ 2 }\quad =\quad a^{ 2 }+b^{ 2 }+2ab\\ a=9,\quad b=0.4\\ =\quad 9^{ 2 }+0.4^{ 2 }+2*9*0.4\\ =\quad 88.36$$
Evaluate using the expansion of $$(a+b)^2$$ or $$(a-b)^2$$ :
$$(92)^2$$
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0%
8444
0%
8464
0%
8474
0%
8414
Explanation
Given: $$ {92}^{2} = {(90 + 2)}^{2} $$
It is in the form of $$ {(a+b)}^{2} $$, where $$ a = 90, b = 2 $$.
Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab $$ we get,
$$ {92}^{2} = {(90 + 2)}^{2} $$
$$= {90}^{2} + { 2 }^{ 2 } + 2\times 90 \times 2 $$
$$= 8100 + 4 + 360 = 8464 $$
Evaluate using expansion of $$(a+b)^2$$ or $$(a-b)^2$$ :
$$(188)^2$$
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0%
35444
0%
35484
0%
35344
0%
35384
Explanation
$$ {188}^{2} = {(200 - 12)}^{2} $$
It is the form of $$ {(a - b)}^{2} $$, where $$ a = 200, b = 12 $$
Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab $$
$$ {188}^{2} = {(200 - 12)}^{2} = {200}^{2} + { 12 }^{ 2 } - 2\times 200 \times 12 = 40000 + 144 - 4800 = 35344 $$
State true or false:
Zero of a polynomial is always $$0$$.
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0%
True
0%
False
Explanation
Zero of a polynomial is not always $$0$$.
e.g. Let $$p(x)=x+3$$
Zero of $$x+3$$ is $$-3$$.
That is, the given statement is false.
Hence, option $$B$$$ is correct.
Find the zeroes of
$$p(x)=x-4$$.
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
Explanation
Given, $$p(x)=x-4$$.
The zero of $$p(x)$$ is given by $$p(x)=0$$
Thus,
$$x-4=0$$
$$=>x=4$$.
Hence, option $$D$$ is correct.
Find the zeroes of the following polynomial:
$$g(x)=2x+1$$.
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0%
$$-\dfrac{1}{2}$$
0%
$$-\dfrac{1}{3}$$
0%
$$-\dfrac{2}{3}$$
0%
$$0$$
Explanation
To find zero of $$g(x)=2x+1$$:
Consider, $$g(x)=0 $$.
$$\implies 2x+1=0$$
$$\implies 2x=-1$$
$$\implies x=\dfrac{-1}2$$.
Therefore, zero of $$g(x)=2x+1$$ is $$\dfrac{-1}2$$.
Hence, option $$A$$ is correct.
The degree of
$$5t-\sqrt 7$$ is
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0%
$$5$$
0%
$$1$$
0%
$$7$$
0%
$$0$$
Explanation
Clearly, the given the given expression $$5t-\sqrt7$$ is a linear polynomial, means the highest power of $$t$$ is 1.
$$\therefore$$ degree of the given polynomial is 1.
State whether True or False, if t
he following are zeros of the polynomial, indicated against them:
$$p(x)=lx+m, \ x=-\dfrac {m}{l}$$.
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0%
True
0%
False
Explanation
Given, $$p(x)=lx+m$$.
On substituting $$x=\dfrac{-m}l$$ in $$p(x)$$, we get,
$$p\left(\dfrac{-m}l\right)=l\times\left(\dfrac{-m}l\right)+m=-m+m=0$$,
Since $$p(x)=0$$ at $$x=\dfrac{-m}l$$,
Therefore, $$x=\dfrac{-m}l$$ is zero of polynomial $$p(x)=lx+m$$.
Hence, it is true and option $$A$$ is correct.
State whether True or False, if t
he following are zeros of the polynomial, indicated against them:
$$p(x)=(x+1)(x-2), \ x=-1, 2$$.
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0%
True
0%
False
Explanation
Given, $$p(x)=(x+1)(x-2)$$.
On substituting $$x=-1$$ in $$p(x)$$, we get,
$$p(-1)=(-1+1)(-1-2)=0$$.
Again on substituting $$x=2$$ in $$p(x)$$, we get,
$$p(2)=(2+1)(2-2)=0$$.
Since $$p(x)=0$$ at $$x=-1,2$$ .
Therefore, $$x=-1,2$$ are zeros of polynomial $$p(x)=(x+1)(x-2)$$.
That is, it is true and option $$A$$ is correct.
The degree of
$$5x^3+4x^2+7x$$ is
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0%
$$1$$
0%
$$2$$
0%
$$4$$
0%
$$3$$
Explanation
$$5x^3+4x^2+7x$$
The degree is the value of the greatest exponent of any expression (except the constant ) in the polynomial.
Here, term with greatest exponent $$= 5x^3$$
exponent of this term $$=3$$
Thus, degree $$=3$$
Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = 0
Report Question
0%
5
0%
4
0%
3
0%
2
Explanation
$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=0$$
$$=0^3+0^2+0+5$$
$$=5$$
State True or False, if t
he following are zeros of the polynomial, indicated against them:
$$p(x)=3x+1, \ x=-\dfrac {1}{3}$$.
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0%
True
0%
False
Explanation
Given, $$p(x)=3x+1$$.
On substituting $$x=\dfrac{-1}3$$ in $$p(x)$$, we get,
$$p\left(\dfrac{-1}3\right)=3\left(\dfrac{-1}3\right)+1=-1+1=0$$.
Since $$p(x)=0$$ at $$x=\dfrac{-1}3$$ ,
therefore, $$x=\dfrac{-1}3$$ is zero of polynomial $$p(x)=3x+1$$.
Hence, it is true and option $$A$$ is correct.
Find the zero of the polynomial $$p(x)=x+5$$.
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0%
$$-5$$
0%
$$5$$
0%
$$0$$
0%
$$3$$
Explanation
Here, $$p(x)=x+5$$.
In order to find zero of $$p(x)$$, we equate $$p(x)$$ to $$0$$.
$$p(x)=0$$
$$x+5=0$$
$$\Rightarrow x=-5$$.
$$\therefore -5$$ is the zero of $$p(x)$$.
Therefore, option $$A$$ is correct.
Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = -1
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0%
0
0%
4
0%
2
0%
6
Explanation
$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=-1$$
$$=(-1^3)+(-1^2)+(-1)+5$$
$$=-1+1-1+5$$
$$=4$$
Substituting $$x= -3$$ in
$$x^2+5x +4$$ gives
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0%
$$-2$$
0%
$$2$$
0%
$$15$$
0%
$$-15$$
Explanation
For $$x=-3$$
$$x^2+5x+4$$
$$=(-3)^2+5(-3)+4$$
$$=9-15+4$$
$$=-2 \neq R.H.S$$
The correct statement is $$x^2+5x+4=-2$$
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