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Introduction to factorization - class-X

Description: introduction to factorization
Number of Questions: 24
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Tags: factorization operations on algebraic expressions algebraic expression algebraic formulae - expansion of squares maths factorisation factors and factorization of algebraic expressions real numbers equations and rearranging formulae straight lines and quadratic equations factorization-1 factors and multiples
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Factorise ${\left( {3 - 4y - 7{y^2}} \right)^2} - {\left( {4y + 1} \right)^2}$ 

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $\left( {4 - 7{y^2}} \right)\left( {2 - 8y - 7{y^2}} \right)$

  2. $\left( {7{y^2} - 4} \right)\left( {2 - 8y - 7{y^2}} \right)$

  3. $\left( {4 - 7{y^2}} \right)\left( {7{y^2} + 8y - 2} \right)$

  4. $\left( {7{y^2} - 4} \right)\left( {7{y^2} - 8y - 2} \right)$

Show answer
Correct Option: A
Explanation:

We have,

${\left( {3 - 4y - 7{y^2}} \right)^2} - {\left( {4y + 1} \right)^2}$ 

We know that
$a^2-b^2=(a+b)(a-b)$

Therefore,
$\Rightarrow (3 - 4y - 7y^2+4y + 1)(3-4y-7y^2-4y-1) $ 

$\Rightarrow (4 - 7y^2)(2-8y-7y^2) $ 

Hence, this is the answer.

Factorise :

$4x+12$

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $4(x+2)$

  2. $4(x+3)$

  3. $3(x+4)$

  4. none of these

Show answer
Correct Option: B
Explanation:

$4x+12=4(x+3)$

Factorize : $y^2 + 100y$
maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $y(y+100)$

  2. $y^2(y+100)$

  3. $y(y^2+100)$

  4. none of these

Show answer
Correct Option: A
Explanation:

We need to find value of $y^2+100y$

Taking $y$ common, we get $y(y+100)$.
Hence, option A is correct.

If (x+1) is a factor of $\displaystyle x^{3}+11x^{2}+15x+a$ then the value of 'a' is

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. 2

  2. 3

  3. 5

  4. 4

Show answer
Correct Option: C
Explanation:

$x+1\div x^{3}+11x^{2}+15x+a\setminus x^{2}+10x+5$

                $x^{3}+x^{2}$ ( subtracted)
               ..............................................................
                               $10x^{2}+15x+a$
                               $10x^{2}+10x$ ( subtracted)
                            ....................................................................
                                          $5x+a$
                                          $5x+5$  ( subtracted)
.....................................................................................................
                                             a-5
If x+1 is the factor of $x^{3}+11x^{2}+15x+a$
Then a-5=0 or a=5

Factorization is the  ......... process of multiplication.

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. equal

  2. same

  3. inverse

  4. common

Show answer
Correct Option: C
Explanation:

Factorization is the inverse process of multiplication.

________ is a method of writing numbers as the product of their factors or divisors.

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. Polynomial

  2. Factorisation

  3. Division algorithm

  4. Quadratic equation

Show answer
Correct Option: B
Explanation:

Factorisation is a method of writing numbers as the product of their factors or divisors.
Example: $4x^2+2x$ is a factor $2x(2x+1)$
By multiplying the factor we get the original number.

If the polynomial $f(x)$ is such that $f(-43) = 0$, which of the following is the factor of $f(x)$?

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $x - 43$

  2. $x$

  3. $x - 7$

  4. $x + 43$

Show answer
Correct Option: D
Explanation:

$f(-43)=0$ 
$\Rightarrow -43$ is root of $f(x)$

$\Rightarrow (x+43).g(x)=f(x)$ for some function $ g(x)$
$\Rightarrow (x+43)$ is the factor of $f(x)$

The denominator of an algebraic fraction should not be

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $1$

  2. $0$

  3. $4$

  4. $7$

Show answer
Correct Option: B
Explanation:

If denominator of algebraic function is zero then it not defined

Hence denominator of any algebraic fraction should not be $0$

If the sum of two integers is $-2$ and their product is $-24$, the numbers are

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $6$ and $4$

  2. $-6$ and $4$

  3. $-6$ and $-4$

  4. $6$ and $-4$

Show answer
Correct Option: B
Explanation:
Let $p$ and $q$ be the required integers

If $p+q$ and $pq$ are known then quadratic equation corresponding to roots as $p$ and $q$ is given by,
$x^2-(p+q)x+pq=0$
$\Rightarrow x^2+2x-24=0$, substitute the given values
$\Rightarrow x^2+6x-4x-24=0$, split the middle term
$\Rightarrow (x^2+6x)+(-4x-24)=0$, group pair of terms
$\Rightarrow x(x+6)-4(x+6)=0$, factor each binomials 
$\Rightarrow (x+6)(x-4)=0$, factor out common factor 
$\Rightarrow x=-6$ or $x=4$, set each factor to $0$

Hence $p$ and $q$ are $-6$ and $4$

The value of  $k$  for which  $x - 1$  is a factor of the polynomial  $4 x ^ { 3 } + 3 x ^ { 2 } - 4 x + k$  is

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $3$

  2. $0$

  3. $1$

  4. $-3$

Show answer
Correct Option: D
Explanation:

$x-1$ is a factor of $4{x^2} + 3{x^2} - 4x + k$


put $x=1$


$4{x^2} + 3{x^2} - 4x + k=0$

$ \Rightarrow 4{\left( 1 \right)^2} + 3{\left( 1 \right)^2} - 4\left( 1 \right) + k = 0$

$ \Rightarrow 4 + 3 - 4 + k = 0$

$ \Rightarrow k =  - 3$

Hence,
option $(D)$ is correct answer.

Factorise : $6xy^2 + 4x^2y$ 

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $2xy(3x+y)$

  2. $xy(3x+2y)$

  3. $2xy(2x+3y)$

  4. none of these

Show answer
Correct Option: C
Explanation:

The common factor between $6xy^2$ and $4x^2y$ is $2xy$ that is the HCF of $6xy^2$ and $4x^2y$ is $2xy$. 

Therefore, we take $2xy$ as a common factor in the expression $6xy^2+4x^2y$ as shown below:
$6xy^2+4x^2y=2xy(2x+3y)$
Hence, the factors of $6xy^2+4x^2y$ are $2xy$ and $(2x+3y)$.

Factorise :
$\displaystyle 121ac-16a^{2}b^{2}$

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $a(121c-16ab^{2})$

  2. $a(121c+16ab^{2})$

  3. $a(121c-16ac^{2})$

  4. none of these

Show answer
Correct Option: A
Explanation:

$121ac-16a^2b^2$

$=121\times a\times c-16\times a\times a\times b\times b$
$=a(121c-16ab^2)$

Which of the following is an example of factorisation?

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $x^2+2x=x(x+2)$

  2. $x^2+2x=x(x+1)$

  3. $x^2+2x=x(x+3)$

  4. None of the above

Show answer
Correct Option: A
Explanation:

The common factor between $x^2$ and $2x$ is $x$ that is the HCF of $x^2$ and $2x$ is $x$. 


Therefore, we take $x$ as a common factor in the expression $x^2+2x$ as shown below:


$x^2+2x=x(x+2)$


Hence, the factorization of $x^2+2x$ is $x(x+2)$.

Factorise : $5mn+15mnp$

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $5mn(1 + 3p)$

  2. $3mn(1 + 5p)$

  3. $5mn(1 - 3p)$

  4. none of these

Show answer
Correct Option: A
Explanation:

The common factor between $5mn$ and $15mnp$ is $5mn$ that is the HCF of $5mn$ and $15mnp$ is $5mn$. 


Therefore, we take $5mn$ as a common factor in the expression $5mn+15mnp$ as shown below:


$5mn+15mnp=5mn(1+3p)$


Hence, the factorization of $5mn+15mnp$ is $5mn(1+3p)$.

Simplify: $\displaystyle \left( -80{ m }^{ 4 }npq \right) \div 10{ m }^{ 3 }{ pqn }^{ 2 }$

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $-8mn$

  2. $-8mnpq$

  3. $-8m$

  4. $\dfrac {-8m}{n}$

Show answer
Correct Option: D
Explanation:

$\displaystyle \left( -80{ m }^{ 4 }npq \right) \div 10{ m }^{ 3 }{ pqn }^{ 2 }=\frac { -80{ m }^{ 4 }npq }{ 10{ m }^{ 3 }{ pqn }^{ 2 } } $

$\displaystyle =\quad -8\times \frac { { m }^{ 4 } }{ { m }^{ 3 } } \times \frac { n }{ { n }^{ 2 } } \times \frac { p }{ p } \times \frac { q }{ q } $
$\displaystyle =\quad -8{ m }^{ 4-3 }\times { n }^{ 1-2 }$
$\displaystyle =\quad -8m\times { n }^{ -1 }$
$\displaystyle =\quad \frac { -8m }{ n } \left( \because { n }^{ -1 }=\frac { 1 }{ n }  \right) $

The factorisation of $ \left (21a^2+3a \right )$ is

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $3a(7a+1)$

  2. $7a(3a+1)$

  3. $3a(7a+3a)$

  4. $3a(a+7)$

Show answer
Correct Option: A
Explanation:

The factorisation of $21a^2+3a$ is $3a(7a+1)$
Taking common terms out, we get $3a(7a+1)$.

Multiplying factors is an example of

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. polynomial

  2. quadratic equation

  3. division algorithm

  4. factorisation

Show answer
Correct Option: D
Explanation:

Multiplying factors is an example of factorisation.
Example: $4x^2+2x$ is a factor $2x(2x+1)$
By multiplying the factor we get $2x(2x+1) = 4x^2+2x$

If $f(x)$ and $g(x)$ are two polynomials with integral coefficients which vanish at $x = \dfrac {1}{2}$, then what is the factor of HCF of $f(x)$ and $g(x)$?

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $x - 1$

  2. $x - 2$

  3. $2x - 1$

  4. $2x + 1$

Show answer
Correct Option: C
Explanation:

Given, $x = \dfrac {1}{2}$

$ \Rightarrow (2x - 1) = 0$
Therefore, $ (2x - 1)$ is satisfying both $f(x)$ and $g(x)$, so $(2x - 1)$ is the factor of H.C.F. of $f(x)$ and $g(x)$.

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx+c=0, a \neq 0$ then the wrong statement is

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $\alpha ^2+\beta ^2=\dfrac{b^2-2ac}{a^2}$

  2. $\alpha \beta =\frac{c}{a}$

  3. $\alpha +\beta =\frac{b}{a}$

  4. $\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{b}{c}$

Show answer
Correct Option: C
Explanation:
We know that sum of roots $= - \dfrac{b}{a}$ and product of roots is  $\dfrac{c}{a}$
Hence option (C) is wrong statement.

Consider the following statements :
1. $x - 2$ is a factor of $x^{3} - 3x^{2} + 4x - 4$
2. $x + 1$ is a factor of $2x^{3} + 4x + 6$
3. $x - 1$ is a factor of $x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x + 1$
Of these statements

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. 1 and 2 are correct

  2. 1, 2 and 3 are correct

  3. 2 and 3 are correct

  4. 1 and 3 are correct

Show answer
Correct Option: A
Explanation:
  1. Remainder $=2^3-3\times 2^2+4\times 2-4$
    $= 8-12+8-4 = 0$
    Hence $x-2$ is a factor.
    2. Remainder$= 2(-1)^3+4(-1)+6$
    $= -2-4+6 = 0$
    Hence $x + 1$ is a factor.
    3. Ramainder $= 1^6-1^5+1^5-1^3+1^2-1+1 = 1$
    Hence $x - 1$ is not a factor.
    $\therefore$ Statements 1 and 2 are correct.

If $4x^{4} -12x^{3}+x^{^{2}}+3ax-b$ is divided by $x^{2}-1$ then a = ____, and b=___

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $5, 4$

  2. $4,9$

  3. $4,5$

  4. $1, -1$

Show answer
Correct Option: C
Explanation:

${ x }^{ 2 }-1=0\quad \Rightarrow x=\pm 1$

Given a polynomial $P(x)={ 4x }^{ 4 }-12{ x }^{ 3 }+{ x }^{ 2 }+3ax-b$
Using remainder theorem,as $P(x)$ is completely divisible by ${ x }^{ 2 }-1$
$\therefore P(\pm 1)=0\ \therefore P(1)=0\ \Rightarrow 4-12+1+3a-b=0\ \Rightarrow 3a-b=7\ \therefore P(-1)=0\ \Rightarrow 4+12+1-3a-b=0\ \Rightarrow 3a+b=17$
By soling both we get
$a=4$   ;$b=5$

$7+3x$ is a factor of $3x^3+7x$.

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

If $7 + 3x$ is a factor of $p\left( x \right) = 3{x^3} + 7x$, then, $p\left( {\frac{{ - 7}}{3}} \right) = 0$.

Compute $p\left( {\frac{{ - 7}}{3}} \right)$ in the given polynomial.

$p\left( {\frac{{ - 7}}{3}} \right) = 3{\left( { - \frac{7}{3}} \right)^3} + 7\left( {\frac{{ - 7}}{3}} \right)$

$ =  - \frac{{343}}{9} - \frac{{49}}{3}$

$ = \frac{{ - 343 - 147}}{9}$

$ =  - \frac{{490}}{9}$

The value of $p\left( {\frac{{ - 7}}{3}} \right)$is not equal to 0.

The given statement is false.

Divide : $\displaystyle \left( 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } \right)$ by $17mp$

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $\displaystyle { m }^{ 2 }p$

  2. $\displaystyle m{ p }^{ 2 }$

  3. $\displaystyle 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

  4. $\displaystyle mp$

Show answer
Correct Option: C
Explanation:

$\displaystyle \frac { 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } }{ 17mp } $

$\displaystyle =\frac { 17mp\left( 3{ m }^{ 2 }p-2m{ p }^{ 2 } \right)  }{ 17mp } $
$\displaystyle = 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$, then the quadratic equation whose roots are $\cfrac{1}{\alpha}$  and $\cfrac{1}{\beta}$ is

maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

  1. $ax^2+bx+c=0$

  2. $bx^2+ax+c=0$

  3. $cx^2+bx+a=0$

  4. $cx^2+ax+c=0$

Show answer
Correct Option: C
Explanation:
The quadratic equation whose roots are  $\dfrac{1}{\alpha }$ & $\dfrac{1}{\beta }$      is  $x^{2}-\left(\dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)x+\dfrac{1}{\alpha \beta }$
 
=> $ x^{2}-\left(\dfrac{\alpha + \beta }{\alpha \beta}\right)x+\dfrac{1}{\alpha \beta } = 0$
 
also we know that $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ as $\alpha,\beta$ are roots of the equation $ax^2+bx+c$
 
=> $ x^{2}-\left(\dfrac{-b }{c}\right)x+\dfrac{a}{c } = 0$
 
=> $ x^{2}+\left(\dfrac{b }{c}\right)x+\dfrac{a}{c } = 0$
 
=>  $  cx^{2}+b x+a = 0 $


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