Class 10: Remainder and Factor Theorem

$\displaystyle \textbf{Question 1: } \text{In each of the following cases, find the remainder when:}$
$\displaystyle \text{(i) }x^4-3x^2+2x+1\text{ is divided by }(x-1)$
$\displaystyle \text{(ii) }x^3+3x^2-12x+4\text{ is divided by }(x-2)$
$\displaystyle \text{(iii) }x^4+1\text{ is divided by }(x+1)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Required remainder }=\text{Value of the polynomial }x^4-3x^2+2x+1\text{ at }x=1$
$\displaystyle \therefore \text{Remainder}=(1)^4-3(1)^2+2(1)+1=1$
$\displaystyle \text{(ii) Required remainder }=\text{Value of the polynomial }x^3+3x^2-12x+4\text{ at }x=2$
$\displaystyle \therefore \text{Remainder}=(2)^3+3(2)^2-12(2)+4=0$
$\displaystyle \text{(iii) Required remainder }=\text{Value of the polynomial }x^4+1\text{ at }x=-1$
$\displaystyle \therefore \text{Remainder}=(-1)^4+1=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Show that:}$
$\displaystyle \text{(i) }(x-2)\text{ is a factor of }5x^2+15x-50$
$\displaystyle \text{(ii) }(3x+2)\text{ is a factor of }3x^2-x-2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) If }(x-2)\text{ is a factor of }5x^2+15x-50,\text{ then the remainder is }0\text{ at }x=2$
$\displaystyle \text{Remainder}=5(2)^2+15(2)-50=20+30-50=0$
$\displaystyle \therefore (x-2)\text{ is a factor of }5x^2+15x-50$
$\displaystyle \text{(ii) If }(3x+2)\text{ is a factor of }3x^2-x-2,\text{ then the remainder is }0\text{ at }x=-\frac{2}{3}$
$\displaystyle \text{Remainder}=3\left(-\frac{2}{3}\right)^2-\left(-\frac{2}{3}\right)-2=\frac{4}{3}+\frac{2}{3}-2=0$
$\displaystyle \therefore (3x+2)\text{ is a factor of }3x^2-x-2$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{Find which of the following is a factor of }2x^3+3x^2-5x-6:$
$\displaystyle \text{(i) }(x+1)\qquad\text{(ii) }(2x-1)\qquad\text{(iii) }(x+2)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=2x^3+3x^2-5x-6$
$\displaystyle \text{(i) For }(x+1),\text{ put }x=-1$
$\displaystyle p(-1)=2(-1)^3+3(-1)^2-5(-1)-6$
$\displaystyle =-2+3+5-6=0$
$\displaystyle \therefore (x+1)\text{ is a factor of }2x^3+3x^2-5x-6$
$\displaystyle \text{(ii) For }(2x-1),\text{ put }x=\frac{1}{2}$
$\displaystyle p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)-6$
$\displaystyle =\frac{1}{4}+\frac{3}{4}-\frac{5}{2}-6=1-\frac{5}{2}-6=-\frac{15}{2}\neq0$
$\displaystyle \therefore (2x-1)\text{ is not a factor of }2x^3+3x^2-5x-6$
$\displaystyle \text{(iii) For }(x+2),\text{ put }x=-2$
$\displaystyle p(-2)=2(-2)^3+3(-2)^2-5(-2)-6$
$\displaystyle =-16+12+10-6=0$
$\displaystyle \therefore (x+2)\text{ is a factor of }2x^3+3x^2-5x-6$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{Find the value of }a\text{ or }k\text{ if:}$
$\displaystyle \text{(i) }(2x+1)\text{ is a factor of }2x^2+ax-3$
$\displaystyle \text{(ii) }(3x-4)\text{ is a factor of }3x^2+2x-k$
$\displaystyle \text{(iii) }(2x+1)\text{ is a factor of }(3k+2)x^3+(k-1)$
$\displaystyle \text{(iv) }(x-2)\text{ is a factor of }2x^5-6x^4-2ax^3+6ax^2+4ax+8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }(2x+1)\text{ is a factor, put }x=-\frac{1}{2}$
$\displaystyle 2\left(-\frac{1}{2}\right)^2+a\left(-\frac{1}{2}\right)-3=0$
$\displaystyle \frac{1}{2}-\frac{a}{2}-3=0$
$\displaystyle 1-a-6=0$
$\displaystyle a=-5$
$\displaystyle \text{(ii) Since }(3x-4)\text{ is a factor, put }x=\frac{4}{3}$
$\displaystyle 3\left(\frac{4}{3}\right)^2+2\left(\frac{4}{3}\right)-k=0$
$\displaystyle \frac{16}{3}+\frac{8}{3}-k=0$
$\displaystyle 8-k=0$
$\displaystyle k=8$
$\displaystyle \text{(iii) Since }(2x+1)\text{ is a factor, put }x=-\frac{1}{2}$
$\displaystyle (3k+2)\left(-\frac{1}{2}\right)^3+(k-1)=0$
$\displaystyle -\frac{3k+2}{8}+k-1=0$
$\displaystyle -3k-2+8k-8=0$
$\displaystyle 5k-10=0$
$\displaystyle k=2$
$\displaystyle \text{(iv) Since }(x-2)\text{ is a factor, put }x=2$
$\displaystyle 2(2)^5-6(2)^4-2a(2)^3+6a(2)^2+4a(2)+8=0$
$\displaystyle 64-96-16a+24a+8a+8=0$
$\displaystyle 16a-24=0$
$\displaystyle a=\frac{24}{16}=\frac{3}{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: } \text{Find the value of }a\text{ and }b\text{, when:}$
$\displaystyle \text{(i) }(x-2)\text{ and }(x+3)\text{ are both factors of }x^3+ax^2+bx-12$
$\displaystyle \text{(ii) }(x-1)\text{ and }(x+2)\text{ are both factors of }x^3+(3a+1)x^2+bx-18$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }(x-2)\text{ is a factor, put }x=2$
$\displaystyle (2)^3+a(2)^2+b(2)-12=0$
$\displaystyle 8+4a+2b-12=0$
$\displaystyle 2a+b=2\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x+3)\text{ is a factor, put }x=-3$
$\displaystyle (-3)^3+a(-3)^2+b(-3)-12=0$
$\displaystyle -27+9a-3b-12=0$
$\displaystyle 3a-b=13\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Adding (1) and (2),}$
$\displaystyle 5a=15$
$\displaystyle a=3$
$\displaystyle \text{Putting }a=3\text{ in (1),}$
$\displaystyle 2(3)+b=2$
$\displaystyle b=-4$
$\displaystyle \therefore a=3\text{ and }b=-4$
$\displaystyle \text{(ii) Since }(x-1)\text{ is a factor, put }x=1$
$\displaystyle (1)^3+(3a+1)(1)^2+b(1)-18=0$
$\displaystyle 1+3a+1+b-18=0$
$\displaystyle 3a+b=16\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x+2)\text{ is a factor, put }x=-2$
$\displaystyle (-2)^3+(3a+1)(-2)^2+b(-2)-18=0$
$\displaystyle -8+4(3a+1)-2b-18=0$
$\displaystyle -8+12a+4-2b-18=0$
$\displaystyle 6a-b=11\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Adding (1) and (2),}$
$\displaystyle 9a=27$
$\displaystyle a=3$
$\displaystyle \text{Putting }a=3\text{ in (1),}$
$\displaystyle 9+b=16$
$\displaystyle b=7$
$\displaystyle \therefore a=3\text{ and }b=7$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: } \text{When }x^3+2x^2-kx+4\text{ is divided by }(x-2),$
$\displaystyle \text{the remainder is }k.\text{ Find }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=x^3+2x^2-kx+4$
$\displaystyle \text{When }x=2,\text{ remainder }=k$
$\displaystyle p(2)=k$
$\displaystyle (2)^3+2(2)^2-k(2)+4=k$
$\displaystyle 8+8-2k+4=k$
$\displaystyle 20-2k=k$
$\displaystyle 3k=20$
$\displaystyle k=\frac{20}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: } \text{Find the value of }a\text{, if the division of }ax^3+9x^2+4x-10$
$\displaystyle \text{by }(x+3)\text{ leaves a remainder of }5.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=ax^3+9x^2+4x-10$
$\displaystyle \text{When }x=-3,\text{ remainder }=5$
$\displaystyle p(-3)=5$
$\displaystyle a(-3)^3+9(-3)^2+4(-3)-10=5$
$\displaystyle -27a+81-12-10=5$
$\displaystyle -27a+59=5$
$\displaystyle -27a=-54$
$\displaystyle a=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: } \text{If }x^3+ax^2+bx+6\text{ has }(x-2)\text{ as a factor}$
$\displaystyle \text{and leaves a remainder of }3\text{ when divided by }(x-3),\text{ find }a\text{ and }b.\hfill\text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=x^3+ax^2+bx+6$
$\displaystyle \text{Since }(x-2)\text{ is a factor, }p(2)=0$
$\displaystyle (2)^3+a(2)^2+b(2)+6=0$
$\displaystyle 8+4a+2b+6=0$
$\displaystyle 4a+2b=-14\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since the remainder is }3\text{ when divided by }(x-3),\ p(3)=3$
$\displaystyle (3)^3+a(3)^2+b(3)+6=3$
$\displaystyle 27+9a+3b+6=3$
$\displaystyle 9a+3b=-30\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Multiplying (1) by }3,\text{ we get}$
$\displaystyle 12a+6b=-42\qquad\ldots\ldots\ldots\text{(3)}$
$\displaystyle \text{Multiplying (2) by }2,\text{ we get}$
$\displaystyle 18a+6b=-60\qquad\ldots\ldots\ldots\text{(4)}$
$\displaystyle \text{Subtracting (3) from (4),}$
$\displaystyle 6a=-18$
$\displaystyle a=-3$
$\displaystyle \text{Putting }a=-3\text{ in (1),}$
$\displaystyle 4(-3)+2b=-14$
$\displaystyle -12+2b=-14$
$\displaystyle 2b=-2$
$\displaystyle b=-1$
$\displaystyle \therefore a=-3\text{ and }b=-1$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{Find the value of }a\text{ and }b,\text{ when }2x^3+ax^2+bx-2$
$\displaystyle \text{leaves remainders }7\text{ and }0\text{ when divided by }(2x-3)\text{ and }(x+2)\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=2x^3+ax^2+bx-2$
$\displaystyle \text{When divided by }(2x-3),\text{ put }x=\frac{3}{2}$
$\displaystyle p\left(\frac{3}{2}\right)=7$
$\displaystyle 2\left(\frac{3}{2}\right)^3+a\left(\frac{3}{2}\right)^2+b\left(\frac{3}{2}\right)-2=7$
$\displaystyle \frac{27}{4}+\frac{9a}{4}+\frac{3b}{2}-2=7$
$\displaystyle 27+9a+6b-8=28$
$\displaystyle 9a+6b=9$
$\displaystyle 3a+2b=3\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{When divided by }(x+2),\text{ put }x=-2$
$\displaystyle p(-2)=0$
$\displaystyle 2(-2)^3+a(-2)^2+b(-2)-2=0$
$\displaystyle -16+4a-2b-2=0$
$\displaystyle 2a-b=-9\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{From (2), }b=2a+9$
$\displaystyle \text{Putting }b=2a+9\text{ in (1),}$
$\displaystyle 3a+2(2a+9)=3$
$\displaystyle 3a+4a+18=3$
$\displaystyle 7a=-15$
$\displaystyle a=-\frac{15}{7}$
$\displaystyle \text{Putting }a=-\frac{15}{7}\text{ in (2),}$
$\displaystyle 2\left(-\frac{15}{7}\right)-b=-9$
$\displaystyle -\frac{30}{7}-b=-9$
$\displaystyle b=\frac{33}{7}$
$\displaystyle \therefore a=-\frac{15}{7}\text{ and }b=\frac{33}{7}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{What number should be added to }3x^3-5x^2+6x,\text{ so that when}$
$\displaystyle \text{it is divided by }(x-3),\text{ the remainder is }8?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be added to }3x^3-5x^2+6x$
$\displaystyle \text{When divided by }(x-3),\text{ put }x=3$
$\displaystyle 3(3)^3-5(3)^2+6(3)+a=8$
$\displaystyle 81-45+18+a=8$
$\displaystyle 54+a=8$
$\displaystyle a=-46$
$\displaystyle \therefore \text{The required number is }-46$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{What number should be subtracted from }x^3+3x^2-8x+14,\text{ so that}$
$\displaystyle \text{when it is divided by }(x-2),\text{ the remainder is }10?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be subtracted from }x^3+3x^2-8x+14$
$\displaystyle \text{When divided by }(x-2),\text{ put }x=2$
$\displaystyle (2)^3+3(2)^2-8(2)+14-a=10$
$\displaystyle 8+12-16+14-a=10$
$\displaystyle 18-a=10$
$\displaystyle a=8$
$\displaystyle \therefore \text{The required number is }8$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: } \text{The polynomials }2x^3-7x^2+ax-6\text{ and }x^3-8x^2+(2a+1)x-16$
$\displaystyle \text{leave the same remainder when divided by }(x-2).\text{ Find the value of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For polynomial }2x^3-7x^2+ax-6,$
$\displaystyle \text{when }x=2,\text{ remainder}=2(2)^3-7(2)^2+a(2)-6$
$\displaystyle =16-28+2a-6=2a-18$
$\displaystyle \text{For polynomial }x^3-8x^2+(2a+1)x-16,$
$\displaystyle \text{when }x=2,\text{ remainder}=(2)^3-8(2)^2+(2a+1)(2)-16$
$\displaystyle =8-32+4a+2-16=4a-38$
$\displaystyle \text{Since the remainders are equal,}$
$\displaystyle 2a-18=4a-38$
$\displaystyle 2a=20$
$\displaystyle a=10$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: } \text{If }(x-2)\text{ is a factor of }2x^3+ax^2+bx-14\text{ and}$
$\displaystyle \text{when the expression is divided by }(x-3),\text{ it leaves a remainder }52.\text{ Find }a\text{ and }b.\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=2x^3+ax^2+bx-14$
$\displaystyle \text{Since }(x-2)\text{ is a factor, }p(2)=0$
$\displaystyle 2(2)^3+a(2)^2+b(2)-14=0$
$\displaystyle 16+4a+2b-14=0$
$\displaystyle 4a+2b=-2$
$\displaystyle 2a+b=-1\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since the remainder is }52\text{ when divided by }(x-3),\ p(3)=52$
$\displaystyle 2(3)^3+a(3)^2+b(3)-14=52$
$\displaystyle 54+9a+3b-14=52$
$\displaystyle 9a+3b=12$
$\displaystyle 3a+b=4\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle a=5$
$\displaystyle \text{Putting }a=5\text{ in (1),}$
$\displaystyle 2(5)+b=-1$
$\displaystyle b=-11$
$\displaystyle \therefore a=5\text{ and }b=-11$
$\displaystyle \\$