Tag: introduction to factorization

Questions Related to introduction to factorization

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

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


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 :

$4x+12$

  1. $4(x+2)$

  2. $4(x+3)$

  3. $3(x+4)$

  4. none of these


Correct Option: B
Explanation:

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

Factorize : $y^2 + 100y$
  1. $y(y+100)$

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

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

  4. none of these


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

  1. 2

  2. 3

  3. 5

  4. 4


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.

  1. equal

  2. same

  3. inverse

  4. common


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.

  1. Polynomial

  2. Factorisation

  3. Division algorithm

  4. Quadratic equation


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 $f(x)$?

  1. $x - 43$

  2. $x$

  3. $x - 7$

  4. $x + 43$


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

  1. $1$

  2. $0$

  3. $4$

  4. $7$


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

  1. $6$ and $4$

  2. $-6$ and $4$

  3. $-6$ and $-4$

  4. $6$ and $-4$


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

  1. $3$

  2. $0$

  3. $1$

  4. $-3$


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.