1. (a) State remainder theorem.[br]Solution:[br]If a polynomial [math] f(x) [/math] is divided by [math] x-a [/math] then the remainder is [math] f(a) [/math].[br][br]1. (b) If [math] ax + b ( a \neq 0 ) [/math] divides [math] f(x),[/math] what is the remainder ?[br]Solution:[br]Here, zero of polynomial [math] ax+b \text{ is } \frac{-b}{a} [/math].[br]Remainder [math] = f\left(\frac{-b}{a} \right) [/math][br][br]2. Use remainder theorem and find the remainder in each of the following:[br](a) [math] (2x^3-5x^2+x-5 ) \div (x-2) [/math] [br]Solution:[br]Let, [math] f(x) = 2x^3-5x^2+x-5 [/math][br]Zero of polynomial [math] x-2 [/math] is 2.[br][math]\begin{align} \text{Remainder } & = f(2) \\ & = 2(2)^3-5(2)^2+2-5 \\ & = 2 \times 8 -5 \times 4 +2 -5 \\ & = 16-20+2-5 \\ & = - 7 \end{align} [/math] [br][br](b) [math] (4x^3+7x^2-3x+2)\div (x+2) [/math][br]Solution:[br]Let, [math] f(x) = 4x^3+7x^2-3x+2 [/math][br]Zero of polynomial [math] x+2 [/math] is [math] -2 [/math][br][math]\begin{align} \text{Remainder } & = f(-2) \\ & = 4(-2)^3+7(-2)^2-3(-2)+2 \\ & = 4 (-8) +7(4)+6+2 \\ & = -32+28+6+2 \\ & = 4 \end{align} [/math] [br][br](c) [math] (x^4-3x^2+15) \div (x-1) [/math][br]Solution:[br]Let, [math] f(x) =x^4-3x^2+15 [/math][br]Zero of polynomial [math] x-1 [/math] is [math] 1 [/math][br][math]\begin{align} \text{Remainder } & = f(1) \\ & = 1^4-3(1)^2+15 \\ & =1-3+15\\ & = 13 \end{align} [/math] [br][br](d) [math] (x^5+x^3+20) \div (2x-1) [/math][br]Solution:[br]Solution:[br]Let, [math] f(x) =x^5+x^3+20 [/math][br]Zero of polynomial [math] 2x-1 [/math] is [math] \frac{1}{2} [/math][br][math]\begin{align} \text{Remainder } & = f\left(\frac{1}{2}\right) \\ & = \left(\frac{1}{2} \right)^5+\left( \frac{1}{2} \right)^3 +20\\ & =\frac{1}{32}+\frac{1}{8} + 20 \\ & = \frac{1+4+ 32\times 20}{32} \\ & = \frac{1+4+640}{32} \\ & =\frac{645}{32} \end{align} [/math] [br] [br](e) [math] 7x^4-6x^3+8x^2-10x+9 \div (3x-9) [/math][br]Solution:[br]Let, [math] f(x) =7x^4-6x^3+8x^2-10x+9 [/math][br]Zero of polynomial [math] 3x-9 [/math] is [math] 3 [/math][br][math]\begin{align} \text{Remainder } & = f(3) \\ & = 7(3)^4-6(3)^3+8(3)^2-10(3)+9 \\ & = 7(81)-6( 27)+8(9) -30+9 \\ & = 567-162+72-30+9 \\ & = 456 \end{align} [/math][br][br](f) [math] 6x^4-4x^3+6x^2+8x+10 \div (2x+3) [/math][br]Solution:[br]Let, [math] f(x) =6x^4-4x^3+6x^2+8x+10 [/math][br]Zero of polynomial [math] 2x+3 [/math] is [math] \frac{-3}{2} [/math][br][math]\begin{align} \text{Remainder } & = f\left( \frac{-3}{2}\right) \\ & = 6\left(\frac{-3}{2} \right)^4 - 4 \left( \frac{-3}{2} \right)^3 +6 \left( \frac{-3}{2} \right)^2 +8\left( \frac{-3}{2} \right) + 10 \\ & = 6 \left( \frac{ 81}{16} \right) -4 \left(\frac{-27 }{8} \right) + 6\left( \frac{9}{4} \right) - 12 + 10 \\ & = \\ & = 3\times \frac{81}{8} + \frac{27}{2}+ 3\times \frac{9}{2} - 2 \\ & = \frac{3\times 81+4\times 27 + 4\times 27 - 2\times 8}{32} \\ & = \frac{243+108+108-16 }{ 8 } \\ & =\frac{443}{8 } \end{align} [/math] [br][br](g) [math] 9x^5-7x^2+12x+10 \div (3x+1) [/math][br]Solution:[br]Let, [math] f(x) = 9x^5-7x^2+12x+10 [/math][br]Zero of polynomial [math] 3x+1 [/math] is [math] \frac{-1}{3} [/math][br][math]\begin{align} \text{Remainder } & = f\left( \frac{-1}{3} \right) \\ & = 9\left(\frac{-1}{3} \right)^5 - 7 \left( \frac{-1}{3} \right)^2 +12 \left( \frac{-1}{3} \right) + 10 \\ & = 9 \left( \frac{ -1}{243} \right) - 7 \left(\frac{1 }{ 9} \right) -4 + 10 \\ & = \\ & = \frac{-1}{27}-\frac{7}{9}+6 \\ & = \frac{-1-7\times 3 + 27 \times 6 }{27} \\ & = \frac{-1-21+162 }{ 27 } \\ & =\frac{140}{27} \end{align} [/math] [br][br]3. (a) If [math] x^4+2x^2-4x+k [/math] is divided by [math] x-2 [/math], the remainder is 4, find the value of [math] k [/math], using remainder theorem.[br]Solution:[br]Let, [math] f(x) = x^4+2x^2-4x+k [/math] [br]Zero of [math] x-2 [/math] is 2.[br][math] \begin{align} & \text{Remainder } = 4 \\ & \text{ or, } f(2) = 4 \\ & \text{ or, } (2)^4+2(2)^2-4(2)+k = 4\\ & \text{ or, } 16+8-8+k = 4 \\ & \text{ or, } 16 + k = 4 \\ & \text{ or, } k = 4 - 16 \\ & \therefore k = - 12 \end{align} [/math] [br][br]3. (b) If [math] x^3-9x^2+(k+1)x-8 [/math] is divided by [math] x-5 [/math], the remainder is 6, find the value of [math] k [/math], using remainder theorem.[br]Solution:[br]Let, [math] f(x) = x^3-9x^2+(k+1)x-8 [/math] [br]Zero of [math] x-5 [/math] is 5.[br][math] \begin{align} & \text{Remainder } = 6 \\ & \text{ or, } f(5 ) = 6 \\ & \text{ or, } 5^3 - 9 (5)^2 +(k+1) (5) - 8 = 6 \\ & \text{ or, } 125 - 225 +5k + 5 - 8 = 6 \\ & \text{ or, } 5k - 97 = 6 \\ & \text{ or, } 5k = 6+97 \\ & \text{ or, } 5k = 103 \\ & \therefore k = \frac{103}{5} \end{align} [/math] [br][br]3. (c) If [math] x^3-ax^2+8x+11 [/math] and [math] 3x^3-ax^2+7ax+13 [/math], both are divided by [math] (x-1) [/math], remainder is same, find the value of [math] a [/math].[br]Solution:[br]Let, [math] f(x) = x^3-ax^2+8x+11 [/math][br]And [math] g(x) = 3x^3-ax^2+7ax+13 [/math][br]Zero of [math] x-1 [/math] is 1.[br]By question the remainder is same,[br]so,[br][math] \begin{align} & f(1) = g(1) \\ & \text{or, } 1^3 - a(1)^2+8(1)+11 = 3(1)^3-a(1)^2+7a(1)+13 \\ & \text{or, } 1 - a + 8 + 11 = 3 - a + 7a +13 \\ & \text{or, } 20 = 7a +16 \\ & \text{or, } 20-16 = 7a \\ & \text{or, } 4 = 7a \\ & \text{or, } 7a = 4 \\ & \therefore a = \frac{4}{7} \end{align} [/math] [br][br]3. (d) If [math] (x-2) [/math] divides the polynomials [math] 4x^3+2x^2+kx+5 [/math] and [math] kx^2+5x+4 [/math] to get the same remainder, find the value of [math] k [/math].[br]Solution:[br]Let, [math] f(x) = 4x^3+2x^2+kx+5 [/math][br]And [math] g(x) = kx^2+5x+4 [/math][br]Zero of [math] x-2 [/math] is 2.[br]By question the remainder is same,[br]so,[br][math] \begin{align} & f(2) = g(2) \\ & \text{or, } 4(2)^3+2(2)^2+k(2)+5 = k(2)^2+5(2)+4 \\ & \text{or, } 4(8)+2(4)+2k+5=k(4)+10+4 \\ & \text{or, } 32+8+2k+5 = 4k + 14 \\ & \text{or, } 45-14 = 4k-2k \\ & \text{or, } 31 = 2k \\ & \text{or, } 2k = 31 \\ & \therefore k = \frac{31}{2} \end{align} [/math] [br][br][br]4. Take a polynomial function. Take any three linear divisors in the form of [math] (x + a), \ \ (ax + b) \text{ and } (ax – b)[/math]. Use remainder theorem and find the remainder.[br]Solution:[br]