The polynomial P(x) = 2x^3 + mx^2-5 leaves the same remainder when divided by (x-1) or (2x + 3). Find the value of m and the remainder.The polynomial also leaves the same remainder also leaves the same remainder when divided by (qx+r), findthe values of q and r.
Accepted Solution
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Answer:m=7Remainder =4 If q=1 then r=3 or r=-1.If q=2 then r=3.They are probably looking for q=1 and r=3 because the other combinations were used earlier in the problem. Step-by-step explanation:Let's assume the remainders left when doing P divided by (x-1) and P divided by (2x+3) is R.By remainder theorem we have that:P(1)=RP(-3/2)=R[tex]P(1)=2(1)^3+m(1)^2-5[/tex][tex]=2+m-5=m-3[/tex][tex]P(\frac{-3}{2})=2(\frac{-3}{2})^3+m(\frac{-3}{2})^2-5[/tex][tex]=2(\frac{-27}{8})+m(\frac{9}{4})-5[/tex][tex]=-\frac{27}{4}+\frac{9m}{4}-5[/tex][tex]=\frac{-27+9m-20}{4}[/tex][tex]=\frac{9m-47}{4}[/tex]Both of these are equal to R.[tex]m-3=R[/tex][tex]\frac{9m-47}{4}=R[/tex]I'm going to substitute second R which is (9m-47)/4 in place of first R.[tex]m-3=\frac{9m-47}{4}[/tex]Multiply both sides by 4:[tex]4(m-3)=9m-47[/tex]Distribute:[tex]4m-12=9m-47[/tex]Subtract 4m on both sides:[tex]-12=5m-47[/tex]Add 47 on both sides:[tex]-12+47=5m[/tex]Simplify left hand side:[tex]35=5m[/tex]Divide both sides by 5:[tex]\frac{35}{5}=m[/tex][tex]7=m[/tex]So the value for m is 7.[tex]P(x)=2x^3+7x^2-5[/tex]What is the remainder when dividing P by (x-1) or (2x+3)?Well recall that we said m-3=R which means r=m-3=7-3=4.So the remainder is 4 when dividing P by (x-1) or (2x+3).Now P divided by (qx+r) will also give the same remainder R=4.So by remainder theorem we have that P(-r/q)=4.Let's plug this in:[tex]P(\frac{-r}{q})=2(\frac{-r}{q})^3+m(\frac{-r}{q})^2-5[/tex]Let x=-r/qThis is equal to 4 so we have this equation:[tex]2u^3+7u^2-5=4[/tex] Subtract 4 on both sides:[tex]2u^3+7u^2-9=0[/tex]I see one obvious solution of 1.I seen this because I see 2+7-9 is 0.u=1 would do that.Let's see if we can find any other real solutions.Dividing:1 | 2 7 0 -9 | 2 9 9 ----------------------- 2 9 9 0This gives us the quadratic equation to solve:[tex]2x^2+9x+9=0[/tex]Compare this to [tex]ax^2+bx+c=0[/tex][tex]a=2[/tex][tex]b=9[/tex][tex]c=9[/tex]Since the coefficient of [tex]x^2[/tex] is not 1, we have to find two numbers that multiply to be [tex]ac[/tex] and add up to be [tex]b[/tex].Those numbers are 6 and 3 because [tex]6(3)=18=ac[/tex] while [tex]6+3=9=b[/tex].So we are going to replace [tex]bx[/tex] or [tex]9x[/tex] with [tex]6x+3x[/tex] then factor by grouping:[tex]2x^2+6x+3x+9=0[/tex][tex](2x^2+6x)+(3x+9)=0[/tex][tex]2x(x+3)+3(x+3)=0[/tex][tex](x+3)(2x+3)=0[/tex]This means x+3=0 or 2x+3=0.We need to solve both of these:x+3=0Subtract 3 on both sides:x=-3----2x+3=0Subtract 3 on both sides:2x=-3Divide both sides by 2:x=-3/2So the solutions to P(x)=4:[tex]x \in \{-3,\frac{-3}{2},1\}[/tex]If x=-3 is a solution then (x+3) is a factor that you can divide P by to get remainder 4.If x=-3/2 is a solution then (2x+3) is a factor that you can divide P by to get remainder 4.If x=1 is a solution then (x-1) is a factor that you can divide P by to get remainder 4.Compare (qx+r) to (x+3); we see one possibility for (q,r)=(1,3).Compare (qx+r) to (2x+3); we see another possibility is (q,r)=(2,3).Compare (qx+r) to (x-1); we see another possibility is (q,r)=(1,-1).