# Exercise 6.1

On the following page we have given the solution of Exercise 6.1 of Mathematics 9 (Science) published by Caravan Book House, Lahore.

Find the H.C.F of the following expressions.
(i) $39x^7y^3z$ and $91x^5y^6 z^7$
(ii) $102xy^2z$, $85x^2yz$ and $187xyz^2$

Solution:

(i) $39x^7y^3z=13\times 3\times x^7 y^3 z$
$91x^5y^6 z^7=13\times 7\times x^5 y^6 z^7$
H.C.F = $13 x^5y^3z$

(ii) $102xy^2z=2\times 3\times 17 xy^2z$
$85x^2yz=3\times 17 x^2 y z$
$187xyz^2 = 11\times 17 xyz^2$
H.C.F= $17 xyz$

Find the H.C.F of the following expressions by factorization.
(i) $x^2+5x+6$, $x^2-4x-12$
(ii) $x^3-27$, $x^2+6x-27$, $2x^2-18$
(iii) $x^3-2x^2+x$, $x^2+2x-3$, $X^2+3x-4$
(iv) $18(x^3-9x^2+8x)$, $24(x^2-3x+2)$
(v) $36(3x^4+5x^3-2x^2)$, $54(27x^4-x)$

Solution:

(i)

\begin{align} x^2+5x+6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}

\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}

H.C.F= $x+2$

(ii)

\begin{align} x^3-27 &=x^3-3^3,\\ &=(x-3)(x^2+3x+9)\end{align}

\begin{align} x^2+6x-27&=x^2+9x-3x-27,\\ &=x(x+9)-3(x+9)\\ &=(x+6)(x-3) \end{align}

\begin{align} 2x^2-18&=2(x^2-9),\\ &=2(x^2-3^2)\\ &=2(x+3)(x-3) \end{align}

H.C.F= $x-3$

(iii)

\begin{align} x^3-2x^2+x &=x(x^2-2x+1),\\ &=x(x^2-x-x+1),\\ &=x(x(x-1)-1(x-1)),\\ &= x(x-1)(x-1)\end{align}

\begin{align} x^2+2x-3&=x^2+3x-x-3,\\ &=x(x+3)-1(x+3)\\ &=(x+3)(x-1) \end{align}

\begin{align} X^2+3x-4&=x^2+4x-x-4,\\ &=x(x+4)-1(x+4)\\ &=(x+4)(x-1) \end{align}

H.C.F= $x-1$

(iv)

\begin{align} 18(x^3-9x^2+8x) &=2 \times 3 \times 3x(x^2-9x+8),\\ &=2 \times 3 \times 3x(x^2-x-8x+8),\\ &=2 \times 3 \times 3x(x(x-1)-8(x-1)),\\ &= 2 \times 3 \times 3x(x-1)(x-8)\end{align}

\begin{align} 24(x^2-3x+2)&=2 \times 2 \times 2 \times 3(x^2-x-2x+2),\\ &=2 \times 2 \times 2 \times 3(x(x-1)-2(x-1)),\\ &=2 \times 2 \times 2 \times 3(x-1)(x-2)\end{align}

H.C.F= $6(x-1)$

(v)

\begin{align} 36(3x^4+5x^3-2x^2) &=4 \times 9 \times 2(3x^2+5x-2),\\ &=4 \times 9 \times 2(3x^2+6x-x-2),\\ &=4 \times 9 \times 2(3x(x+2)-1(x+2)),\\ &= 4 \times 9 \times 2(x+2)(3x-1)\end{align}

\begin{align} 54(27x^4-x)&=2 \times 3 \times 9x(27x^3-1),\\ &=2 \times 3 \times 9x((3x)^3-(1)^3),\\ &=2 \times 3 \times 9x(3x-1)(9x^2+3x+1)\end{align}

H.C.F= $18(3x-1)$

Find the L.C.M. of the following expressions by factorization.
(i) $39x^7y^3z$, $91x^5y^6z^7$
(ii)$102xy^2z$, $85x^2yz$ , $187xyz^2$

Solution:

(i)

\begin{align} 39x^7y^3z &=3 \times 13 x^7y^3z,\\ &=3 \times 13x^5x^2y^3z\end{align}
\begin{align} 91x^5y^6z^7 &=7 \times 13 x^5y^6z^7,\\ &=7 \times 13x^5y^2y^3zz^6\end{align}

\begin{align} L.C.M.&= (13x^5y^3z)(21x^2y^3z^6)\\ &=273 x^(5+2)y^(3+3)z^(1+6)\\ &=273 x^7y^6z^7 \end{align}

(ii)

\begin{align} 102xy^2z &=2 \times 3 \times 17 xyyz\end{align}
\begin{align} 85x^2yz &=5 \times 17 xxyz\end{align}
\begin{align} 187xyz^2 &=11 \times 17 xyzz\end{align}

\begin{align} L.C.M.&= (17xyz)(2 \times 5 \times 11xyz)\\ &=5610 x^2y^2z^2\end{align}

Find the L.C.M. of the following expressions by factorization.
(i) $x^2-25x+100$, $x^2-x-20$
(ii) $x^2+4x+4$, $x^2-4$,$2x^2+x-6$
(iii) $2(x^4-y^4)$,$3(x^3+2x^2-xy^2-2y^3)$
(iv) $4(x^4-1)$, $6(x^3-x^2-x+1)$

Solution:

(i)

\begin{align} x^2-25x+100 &=(x^2-5x-20x+100)\\ &=(x(x-5)-20(x-5))\\ &= (x-5)(x-20)\end{align}

\begin{align} x^2-x-20&=x^2-5x+4x-20\\ &=x(x-5)+4(x-5)\\ &=(x-5)(x+4) \end{align}

L.C.M.= $(x-5)(x-20)(x+4)$

(ii)

\begin{align} x^2+4x+4 &=(x^2+2x+2x+4)\\ &=(x(x+2)+2(x+2))\\ &= (x+2)(x+2)\end{align}

\begin{align} x^2-4 &=(x-2)(x+2) \end{align}

\begin{align} 2x^2+x-6 &=(2x^2+4x-3x-6)\\ &=(2x(x+2)-3(x+2))\\ &= (x+2)(2x-3)\end{align}

L.C.M.= $(x+2)^2(x-2)(2x-3)$

(iii)

\begin{align} 2(x^4-y^4)&=2((x^2)^2-(y^2)^2)\\ &=2((x^2-y^2)(x^2+y^2))\\ &= 2(x-y)(x+y)(x^2+y^2)\end{align}

\begin{align} 3(x^3+2x^2-xy^2-2y^3)&=3(x^2(x+2y)-y^2(x+2y))\\ &=3(x+2y)(x^2-y^2)\\ &=3(x+2y)(x-y)(x+y) \end{align}

\begin{align} L.C.M.&= 2 \times 3 (x+y)(x-y)(x+2y)(x^2+y^2)\\ &=6 (x^4-y^4)(x+2y)\end{align}

(iv)

\begin{align} 4(x^4-1)&=2 \times 2((x^2)^2-(1^2)^2)\\ &=2 \times 2((x^2-1)(x^2+1))\\ &= 2 \times 2(x-1)(x+1)(x^2+1)\end{align}

\begin{align} 6(x^3-x^2-x+1)&=2 \times 3(x^2(x-1)-1(x-1))\\ &=2 \times 3(x-1)(x^2-1)\\ &=2 \times 3(x-1)^2(x+1) \end{align}

\begin{align} L.C.M.&= 2 \times 2 \times 3 (x+1)(x-1)(x-1)(x^2+1)\\ &=12 (x^4-1)(x-1)\end{align}

For what value of $k$ is $(x+4)$ the H.C.F. of $x^2+x-(2k+2)$ and $2x^2+kx-12$ ?

Solution:

$(x+4)$ will divide completely $x^2+x-(2k+2)$ and $2x^2+kx-12$.

\begin{align} p(x)&=x^2+x-2k-2 \end{align}

put $x=-4$

\begin{align} p(-4)&=(-4)^2+(-4)-2k-2 \\&= 16-4-2k-2\\&=10-2k\\&=R \end{align}

$R$ must be zero.

therefore \begin{align}10-2k &=0 \end{align}

\begin{align}-2k &=-10\end{align}

\begin{align}k &=5\end{align}

\begin{align} q(x) &=2x^2+kx-12 \end{align}

put \begin{align} x &=5 \end{align}

\begin{align} q(-4) &=2(-4)^2-4k-12 \\&=32-4k-12\\&=20-4k\\&=R\end{align}

$R$ must be zero.

\begin{align} 20 -4k &= 0 \end{align}

\begin{align} -4k &= -20 \end{align}

\begin{align} k &= 5 \end{align}

If $(x+3)(x-2)$ is the H.C.F. of $p(x)=(x+3)(2x^2-3x+k)$ and $q(x)=(x-2)(3x^2+7x-l)$, Find $k$ and $l$.

Solution:

$(x+3)(x-2)$ will divide completely $p(x)$ and $q(x)$.

\begin{align} p(x)&=(x+3)(2x^2-3x+k) \end{align}

put $x=2$

\begin{align} p(2)&= (2+3)(2(2)^2-3(2)+k)\\&= 5(8-6+k)\\&=5(2+k)\\&=R \end{align}

$R$ must be zero.

therefore \begin{align}5(2+k) &=0 \end{align}

\begin{align}2+k &=0\end{align}

\begin{align}k &=-2\end{align}

\begin{align} q(x) &=(x-2)(3x^2+7x-l) \end{align}

$(x+3)(x-2)$ will divide completely $q(x)$.

put \begin{align} x &=-3 \end{align}

\begin{align} q(-3) &=(-3-2)(3(-3)^2+7(-3)-l) \\&=(-5)(27-21-2l)\\&=(-5)(6-l)\\&=R\end{align}

$R$ must be zero.

\begin{align} (-5)(6-l) &= 0 \end{align}

\begin{align} 6-l &= 0 \end{align}

\begin{align} l &= 6 \end{align}

The L.C.M. and H.C.F. of two polynomials $p(x)$ and $2(x^4-1)$ and $(x+1)(x^2+1)$ respectively. If $p(x)= x^3+x^2+x+1$, find $q(x)$ are

Solution:

We know $P(x) \times q(x)= L.C.M \times H.C.F.$

\begin{align} x^3+x^2+x+1 \times q(x)&=2(x^4-1)(x+1)(x^2+1)\end{align}

\begin{align} q(x)&=\frac{2(x^4-1)(x+1)(x^2+1)}{x^3+x^2+x+1}\\&=\frac{2(x^4-1)(x+1)(x^2+1)}{x^2(x+1)+1(x+1)}\\&=\frac{2(x^4-1)(x+1)(x^2+1)}{(x+1)(x^2+1)}\\&=2(x^4-1) \end{align}

Let $p(x)= 10(x^2-9)(x^2-3x+2)$ and $q(x)= 10x(x+3)(x-1)^2$. If the H.C.F. of $p(x)$ , $q(x)$ is $10(x+3)(x-1)$, find their L.C.M.

Solution:

We know $P(x) \times q(x)= L.C.M \times H.C.F.$

\begin{align} 10(x+3)(x-1) \times q(x)&=10(x^2-9)(x^2-3x+2)10x(x+3)(x-1)^2\end{align}

\begin{align} q(x)&=\frac{10(x^2-9)(x^2-3x+2)10x(x+3)(x-1)^2}{10(x+3)(x-1)}\\&=(x^2-9)(x^2-3x+2)10x(x-1)\\&=10x(x^2-9)(x-1)(x^2-3x+2)\\&=10x(x^2-9)(x-1)(x^2-2x-x+2)\\&=10x(x^2-9)(x-1)(x(x-2)-1(x-2))\\&=10x(x^2-9)(x-1)(x-2)(x-1)\\&=10x(x^2-9)(x-1)^2(x-2)\end{align}

Let the product of L.C.M. and H.C.F. of two polynomials be $(x+3)^2(x-2)(x+5)$. If one polynomial is $(x+3)(x-2)$ and the second polynomial is $x^2+kx+15$, find the value of k.

Solution:

We know L.C.M. \times H.C.F = (x+3)^2(x-2)(x+5)

\begin{align} p(x)&=(x+3)(x-2)\end{align}

\begin{align} q(x)&=x^2+kx+15 \end{align}

$P(x) \times q(x)= L.C.M \times H.C.F.$

\begin{align} (x^2+kx+15)(x+3)(x-2)&= (x+3)^2(x-2)(x+5)\end{align}

\begin{align} (x^2+kx+15)&= \frac{(x+3)^2(x-2)(x+5)}{(x+3)(x-2)}\end{align}

\begin{align} (x^2+kx+15)&= (x+3)(x+5)\end{align}

\begin{align} x^2+kx+15&= x^2+8x+15\end{align}

\begin{align} kx&= x^2+8x+15-x^2-15\end{align}

\begin{align} kx&= 8x\end{align}

\begin{align} k&= 8\end{align}