H.C.F. and L.C.M. of Polynomials Questions in English

(B) Let $p(x) = 3 + 13x - 30x^2$.
Rearranging the terms: $p(x) = -30x^2 + 13x + 3$.
Factorizing $p(x) = -30x^2 + 18x - 5x + 3 = -6x(5x - 3) - 1(5x - 3) = -(6x + 1)(5x - 3)$.
Let $q(x) = 25x^2 - 30x + 9$.
This is a perfect square trinomial: $q(x) = (5x)^2 - 2(5x)(3) + 3^2 = (5x - 3)^2$.
The $G.C.D.$ (Greatest Common Divisor) is the highest common factor between $p(x)$ and $q(x)$.
Comparing $p(x) = -(6x + 1)(5x - 3)$ and $q(x) = (5x - 3)(5x - 3)$,the common factor is $(5x - 3)$.
Therefore,the $G.C.D.$ is $(5x - 3)$.

(C) Let the two polynomials be $p(x) = (x+3)^{2}(x-2)(x+1)^{2}$ and $q(x) = (x+1)^{3}(x+3)(x+4)$.
The $L.C.M.$ of polynomials is the product of the highest powers of all the distinct factors present in the expressions.
The distinct factors are $(x+3)$,$(x-2)$,$(x+1)$,and $(x+4)$.
The highest power of $(x+3)$ is $(x+3)^{2}$.
The highest power of $(x-2)$ is $(x-2)^{1}$.
The highest power of $(x+1)$ is $(x+1)^{3}$.
The highest power of $(x+4)$ is $(x+4)^{1}$.
Therefore,the $L.C.M.$ is $(x+3)^{2}(x+1)^{3}(x-2)(x+4)$.

(A) Given polynomials are $p(x) = 2x^{2}-3x-2$ and $q(x) = x^{3}-4x^{2}+4x$.
First,factorize $p(x)$:
$p(x) = 2x^{2}-4x+x-2 = 2x(x-2)+1(x-2) = (2x+1)(x-2)$.
Next,factorize $q(x)$:
$q(x) = x(x^{2}-4x+4) = x(x-2)^{2}$.
The $L.C.M.$ is the product of the highest powers of all factors present in the expressions:
$L.C.M. = x(x-2)^{2}(2x+1)$.

(B) First,factorize the given expressions:
$p(x) = 8(x^{3}-x^{2}+x) = 2^{3} \cdot x \cdot (x^{2}-x+1)$
$q(x) = 28(x^{3}+1) = 2^{2} \cdot 7 \cdot (x+1)(x^{2}-x+1)$
Next,identify the common factors:
The numerical $G.C.D.$ of $8$ and $28$ is $2^{2} = 4$.
The common polynomial factor is $(x^{2}-x+1)$.
Therefore,the required $G.C.D.$ is $4(x^{2}-x+1)$.

(A) Let the expressions be $E_1 = 4x^4 + y^4$,$E_2 = 2x^3 - xy^2 - y^3$,and $E_3 = 2x^2 + 2xy + y^2$.
For $E_1 = 4x^4 + y^4$,we can complete the square: $4x^4 + y^4 = (2x^2 + y^2)^2 - 4x^2y^2 = (2x^2 + y^2 - 2xy)(2x^2 + y^2 + 2xy)$.
For $E_2 = 2x^3 - xy^2 - y^3$,we can rewrite it as $2x^3 - 2y^3 + y^3 - xy^2 - y^3 = 2(x^3 - y^3) - y^2(x - y) = 2(x - y)(x^2 + xy + y^2) - y^2(x - y) = (x - y)(2x^2 + 2xy + 2y^2 - y^2) = (x - y)(2x^2 + 2xy + y^2)$.
For $E_3 = 2x^2 + 2xy + y^2$,it is already in its simplest form.
The common factor present in all three expressions is $(2x^2 + 2xy + y^2)$.
Therefore,the $G.C.D.$ is $2x^2 + 2xy + y^2$.

(C) Let the two polynomials be $p(x) = (x+4)^{2}(x-3)^{2}$ and $q(x) = (x-1)(x+4)(x-3)^{2}$.
The $G.C.D.$ (Greatest Common Divisor) is the product of the lowest powers of all common factors present in the given polynomials.
Factors of $p(x)$ are $(x+4)^{2}$ and $(x-3)^{2}$.
Factors of $q(x)$ are $(x-1)$,$(x+4)$,and $(x-3)^{2}$.
The common factors are $(x+4)$ and $(x-3)^{2}$.
The lowest power of $(x+4)$ is $(x+4)^{1}$.
The lowest power of $(x-3)$ is $(x-3)^{2}$.
Therefore,the $G.C.D.$ is $(x+4)(x-3)^{2}$.

(A) Given polynomials are $p(x) = 16 - 4x^{2}$ and $q(x) = x^{2} + x - 6$.
Step $1$: Factorize $p(x)$.
$p(x) = 16 - 4x^{2} = 4(4 - x^{2}) = 4(2 - x)(2 + x) = -4(x - 2)(x + 2)$.
Step $2$: Factorize $q(x)$.
$q(x) = x^{2} + x - 6 = x^{2} + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2)$.
Step $3$: Find the $L.C.M.$
The $L.C.M.$ is the product of the highest powers of all prime factors present in the expressions.
$L.C.M. = -4(x - 2)(x + 2)(x + 3)$.
Simplifying the expression:
$L.C.M. = -4(x^{2} - 4)(x + 3)$.

(B) Let $p(x) = x^{2}-4$.
Using the identity $a^{2}-b^{2} = (a-b)(a+b)$,we get:
$p(x) = (x-2)(x+2)$.
Now,let $q(x) = x^{2}-5x+6$.
To factorize this quadratic,we look for two numbers whose product is $6$ and sum is $-5$. These numbers are $-2$ and $-3$.
$q(x) = x^{2}-2x-3x+6$
$q(x) = x(x-2)-3(x-2)$
$q(x) = (x-2)(x-3)$.
The $G.C.D.$ (Greatest Common Divisor) is the highest degree common factor present in both polynomials.
Comparing $p(x) = (x-2)(x+2)$ and $q(x) = (x-2)(x-3)$,the common factor is $(x-2)$.
Therefore,the $G.C.D.$ is $x-2$.

(A) We know that for any two polynomials $p(y)$ and $q(y)$,the relationship is: $p(y) \times q(y) = H.C.F. \times L.C.M.$
Given:
$H.C.F. = (y-7)$
$L.C.M. = y^{3}-10y^{2}+11y+70$
$p(y) = y^{2}-5y-14 = (y-7)(y+2)$
Substituting the values into the formula:
$(y-7)(y+2) \times q(y) = (y-7) \times (y^{3}-10y^{2}+11y+70)$
$q(y) = \frac{(y-7)(y^{3}-10y^{2}+11y+70)}{(y-7)(y+2)}$
$q(y) = \frac{y^{3}-10y^{2}+11y+70}{y+2}$
Performing polynomial division of $(y^{3}-10y^{2}+11y+70)$ by $(y+2)$:
$y^{3}+2y^{2} - 12y^{2}-24y + 35y+70 = y^{2}(y+2) - 12y(y+2) + 35(y+2)$
$= (y^{2}-12y+35)(y+2)$
Thus,$q(y) = y^{2}-12y+35$.

(C) The $G.C.D.$ (Greatest Common Divisor) is given as $(x-4)$.
Let $p(x) = x^{2}-x-12$. Factoring this,we get $p(x) = (x-4)(x+3)$.
Let $q(x) = x^{2}-mx-8$.
Since $(x-4)$ is the $G.C.D.$,it must be a factor of $q(x)$.
According to the Factor Theorem,if $(x-4)$ is a factor of $q(x)$,then $q(4) = 0$.
Substituting $x = 4$ into $q(x)$:
$q(4) = (4)^{2} - m(4) - 8 = 0$
$16 - 4m - 8 = 0$
$8 - 4m = 0$
$4m = 8$
$m = 2$.

(C) Let the two polynomials be $p(x) = (x-2)^{2}(x+3)(x-4)$ and $q(x) = (x-2)(x+2)(x-5)$.
To find the $G.C.D.$ (Greatest Common Divisor),we identify the common factors present in both expressions.
The factors of $p(x)$ are $(x-2)$,$(x-2)$,$(x+3)$,and $(x-4)$.
The factors of $q(x)$ are $(x-2)$,$(x+2)$,and $(x-5)$.
The only common factor present in both polynomials is $(x-2)$.
Therefore,the $G.C.D.$ of the given polynomials is $(x-2)$.

(B) Let $p(x) = x^{2}-2x-24$ and $q(x) = x^{2}-ax-6$.
Since $(x-6)$ is the $G.C.D.$ of $p(x)$ and $q(x),$ it must be a factor of both polynomials.
According to the Factor Theorem,if $(x-6)$ is a factor of $q(x),$ then $q(6) = 0$.
Substituting $x = 6$ into $q(x)$:
$q(6) = (6)^{2} - a(6) - 6 = 0$
$36 - 6a - 6 = 0$
$30 - 6a = 0$
$6a = 30$
$a = 5$
Thus,the value of $a$ is $5$.

(C) We know that for any two polynomials $p(x)$ and $q(x)$:
$p(x) \times q(x) = L.C.M. \times H.C.F.$
Given:
$p(x) = 4 x^{2}(x^{2}-a^{2})$
$L.C.M. = 36 x^{3}(x+a)(x^{3}-a^{3})$
$H.C.F. = x^{2}(x-a)$
Substituting these values into the formula:
$4 x^{2}(x^{2}-a^{2}) \times q(x) = 36 x^{3}(x+a)(x^{3}-a^{3}) \times x^{2}(x-a)$
Since $(x^{2}-a^{2}) = (x-a)(x+a)$,we can write:
$4 x^{2}(x-a)(x+a) \times q(x) = 36 x^{5}(x+a)(x-a)(x^{3}-a^{3})$
$q(x) = \frac{36 x^{5}(x+a)(x-a)(x^{3}-a^{3})}{4 x^{2}(x+a)(x-a)}$
$q(x) = 9 x^{3}(x^{3}-a^{3})$

(A) Let $p(x) = x^{2}-x-6$ and $q(x) = x^{2}+3x-18$.
Since $(x-a)$ is the $G.C.D.$ of $p(x)$ and $q(x)$,$(x-a)$ must be a common factor of both polynomials.
Factorizing $p(x)$:
$p(x) = x^{2}-3x+2x-6 = x(x-3)+2(x-3) = (x-3)(x+2)$.
Factorizing $q(x)$:
$q(x) = x^{2}+6x-3x-18 = x(x+6)-3(x+6) = (x-3)(x+6)$.
The common factor is $(x-3)$.
Comparing $(x-3)$ with $(x-a)$,we get $a = 3$.

(A) We know that for any two polynomials $p(x)$ and $q(x)$,the product of the polynomials is equal to the product of their $H.C.F.$ and $L.C.M.$
$p(x) \times q(x) = H.C.F. \times L.C.M.$
Given:
$H.C.F. = x(x+a)$
$L.C.M. = 12x^2(x+a)(x^2-a^2)$
$p(x) = 4x(x+a)^2$
Substituting the values into the formula:
$q(x) = \frac{H.C.F. \times L.C.M.}{p(x)}$
$q(x) = \frac{x(x+a) \times 12x^2(x+a)(x^2-a^2)}{4x(x+a)^2}$
$q(x) = \frac{12x^3(x+a)^2(x^2-a^2)}{4x(x+a)^2}$
$q(x) = 3x^2(x^2-a^2)$

(B) Let $p(x) = 8(x^{4}-16)$.
Factorizing $p(x)$:
$p(x) = 8(x^{2}-4)(x^{2}+4) = 8(x-2)(x+2)(x^{2}+4) = 4 \times 2(x-2)(x+2)(x^{2}+4)$.
Let $q(x) = 12(x^{3}-8)$.
Factorizing $q(x)$ using the formula $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$:
$q(x) = 12(x-2)(x^{2}+2x+4) = 4 \times 3(x-2)(x^{2}+2x+4)$.
The $G.C.D.$ is the product of the common factors with the lowest power.
The common factor is $4(x-2)$.
Therefore,the $G.C.D.$ is $4(x-2)$.

(A) Let $p(x) = (x+3)(-6x^2+5x+4)$.
Factorizing the quadratic expression: $-6x^2+5x+4 = -6x^2+8x-3x+4 = -2x(3x-4)-1(3x-4) = -(3x-4)(2x+1)$.
So,$p(x) = -(x+3)(3x-4)(2x+1)$.
Let $q(x) = (2x^2+7x+3)(x+3)$.
Factorizing the quadratic expression: $2x^2+7x+3 = 2x^2+6x+x+3 = 2x(x+3)+1(x+3) = (2x+1)(x+3)$.
So,$q(x) = (2x+1)(x+3)(x+3) = (x+3)^2(2x+1)$.
The $L.C.M.$ is the product of the highest powers of all prime factors present in the expressions.
$L.C.M. = -(x+3)^2(3x-4)(2x+1)$.

(C) Let $p(x) = 36 x^{2}-49$.
Using the identity $a^{2}-b^{2} = (a-b)(a+b)$,we get:
$p(x) = (6 x)^{2}-(7)^{2} = (6 x+7)(6 x-7)$.
Let $q(x) = 6 x^{2}-25 x+21$.
To factorize this quadratic,we split the middle term:
$q(x) = 6 x^{2}-18 x-7 x+21$
$q(x) = 6 x(x-3)-7(x-3)$
$q(x) = (6 x-7)(x-3)$.
The $G.C.D.$ (Greatest Common Divisor) is the common factor present in both polynomials.
Therefore,$G.C.D. = (6 x-7)$.

(B) First,factorize the first polynomial: $30 x^{2}+13 x-3 = 30 x^{2}+18 x-5 x-3 = 6 x(5 x+3)-1(5 x+3) = (5 x+3)(6 x-1)$.
Next,factorize the second polynomial: $25 x^{2}-30 x+9 = 25 x^{2}-15 x-15 x+9 = 5 x(5 x-3)-3(5 x-3) = (5 x-3)^{2}$.
The $L.C.M.$ is the product of the highest powers of all prime factors present in the expressions.
Therefore,$L.C.M. = (5 x-3)^{2}(5 x+3)(6 x-1)$.

(C) Let the first polynomial be $p(x) = 6 x^{2} + 11 x + 3$.
By splitting the middle term: $6 x^{2} + 9 x + 2 x + 3 = 3 x(2 x + 3) + 1(2 x + 3) = (2 x + 3)(3 x + 1)$.
Let the second polynomial be $q(x) = 2 x^{2} + x - 3$.
By splitting the middle term: $2 x^{2} + 3 x - 2 x - 3 = x(2 x + 3) - 1(2 x + 3) = (2 x + 3)(x - 1)$.
The $G.C.D.$ is the common factor between the two polynomials.
Therefore,$G.C.D. = (2 x + 3)$.

(C) The relationship between two expressions,their $H.C.F.$,and their $L.C.M.$ is given by the formula:
$L.C.M. \times H.C.F. = \text{Product of the two expressions}$
Given that the $H.C.F.$ of $p$ and $q$ is $1$,we substitute this into the formula:
$L.C.M. \times 1 = p \times q$
Therefore,$L.C.M. = p \cdot q$.

(D) First,factorize each expression:
$2x^2 - 4x = 2x(x - 2)$
$3x^4 - 12x^2 = 3x^2(x^2 - 4) = 3x^2(x - 2)(x + 2)$
$2x^5 - 2x^4 - 4x^3 = 2x^3(x^2 - x - 2) = 2x^3(x - 2)(x + 1)$
The common factors present in all three expressions are $x$ and $(x - 2)$.
Therefore,the $H.C.F.$ is $x(x - 2)$.

(D) We know the fundamental relationship between the product of two expressions and their $H.C.F.$ and $L.C.M.$ is given by:
Product of two expressions $= H.C.F. \times L.C.M.$
Given:
Product $= (x+y+z) p^{3}$
$H.C.F. = p^{2}$
Therefore,$L.C.M. = \frac{\text{Product}}{H.C.F.}$
$L.C.M. = \frac{(x+y+z) p^{3}}{p^{2}}$
$L.C.M. = (x+y+z) p$

(A) Since $(x-1)$ is the $H.C.F.$ of the given expressions,it must be a factor of both expressions.
Therefore,substituting $x=1$ into the expressions must result in zero.
For the second expression,$[p x^{2}-q(x+1)]$,we set $x=1$:
$p(1)^{2} - q(1+1) = 0$
$p(1) - q(2) = 0$
$p - 2q = 0$
$p = 2q$.

(B) First,factorize each expression:
$1. x^{2}-y^{2} = (x-y)(x+y)$
$2. x^{3}-y^{3} = (x-y)(x^{2}+xy+y^{2})$
$3. x^{3}-x^{2}y-xy^{2}+y^{3} = x^{2}(x-y)-y^{2}(x-y) = (x^{2}-y^{2})(x-y) = (x+y)(x-y)(x-y) = (x+y)(x-y)^{2}$
To find the $L.C.M.$,take the highest power of each distinct factor present in the expressions:
Factors are $(x+y)$,$(x-y)$,and $(x^{2}+xy+y^{2})$.
The highest power of $(x+y)$ is $(x+y)^{1}$.
The highest power of $(x-y)$ is $(x-y)^{2}$.
The highest power of $(x^{2}+xy+y^{2})$ is $(x^{2}+xy+y^{2})^{1}$.
Therefore,$L.C.M. = (x+y)(x-y)^{2}(x^{2}+xy+y^{2})$.