answer this using polynomials
(2x⁴+x³-19x²+18x+5)÷(2x-5)
Answer (1)
Divide [tex]\(P(x)=2x^4+x^3-19x^2+18x+5\) by \(D(x)=2x-5\).[/tex]1. Leading term: [tex]\(\dfrac{2x^4}{2x}=x^3\).[/tex] Multiply–subtract: [tex]\((2x^4+x^3)- (x^3(2x-5))=(2x^4+x^3)-(2x^4-5x^3)=6x^3\).[/tex] Bring down: [tex]\(6x^3-19x^2\).[/tex]2. Next: [tex]\(\dfrac{6x^3}{2x}=3x^2\).[/tex] Multiply–subtract: [tex]\((6x^3-19x^2)-(3x^2(2x-5))=(6x^3-19x^2)-(6x^3-15x^2)=-4x^2\). Bring down: \(-4x^2+18x\)[/tex].3. Next: [tex]\(\dfrac{-4x^2}{2x}=-2x\)[/tex] Multiply–subtract: [tex]\((-4x^2+18x)-(-2x(2x-5))=(-4x^2+18x)-(-4x^2+10x)=8x\). Bring down: \(8x+5\)[/tex]4. Next: [tex]\(\dfrac{8x}{2x}=4\).[/tex] Multiply–subtract: [tex]\((8x+5)-(4(2x-5))=(8x+5)-(8x-20)=25\)[/tex].Quotient and remainder:[tex]- \(Q(x)=x^3+3x^2-2x+4\)- \(R=25\)[/tex]So,[tex]\[\frac{2x^4+x^3-19x^2+18x+5}{2x-5}= x^3+3x^2-2x+4+\frac{25}{2x-5}.\][/tex]Quick check (Remainder Theorem): [tex]\(R=P\!\left(\tfrac{5}{2}\right)=25\)[/tex], which matches.
