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Solve(2x4+3xcube-2xsquare +4x)+(x+3) by synthetic division method |
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Answer» ong>Answer: Step-by-step explanation:A polynomial p(x) is defined as ⇒p(x)=g(x)Q(x)+r(x)
where g(x)= divisor ; q(x)= quotient and r(x)= remainder
∴ p(x) can be FOUND by multiplying g(x) with q(x) & adding r(x) to the product.
(i).g(x)=(x−2); q(x)=x 2 −x+1; r(x)=4 ∴p(x)=(x−2)[x 2 −x+1]+4 =x 3 −x 2 +x−2x 2 +2x−2+4 =x 3 −3x 2 +3x+2
(II).g(x)=(x+3); q(x)=2x 2 +x+5; r(x)=3x+1 ∴p(x)=(x+3)[2x 2 +x+5]+(3x+1) =2x 3 +x 2 +5x+6x 2 +3x+15+3x+1 =2x 3 +7X 2 +11x+16
(iii).g(x)=(2x+1); q(x)=x 3 +3x 2 −x+1; r(x)=0 ∴p(x)=(2x+1)[x 3 +3x 2 −x+1]+(0) =2x 4 +6x 3 −2x 2 +2x+x 3 +3x 2 −x+1 =2x 4 +7x 3 +x 2 +x+1
(iv).g(x)=(x−1); q(x)=x 3 −x 2 −x−1; r(x)=2x−4 ∴p(x)=(x−1)[x 3 −x 2 −x−1]+(2x−4) =x 4 −x 3 −x 2 −x−x 3 +x 2 +x+1+2x−4 =x 4 −2x 3 +2x−3
(v).g(x)=(x 2 +2x+1); q(x)=x 4 −2x 2 +5x−7; r(x)=4X+12 ∴p(x)=(x 2 +2x+1)[x 4 −2x 2 +5x−7]+(4x+12) =x 6 −2x 4 +5x 3 −7x 2 +2x 5 +4x 3 +10x 2 −14x+x 4 −2x 2 +5x−7+4x+12 =x 6 −x 4 +x 3 +x 2 +2x 5 −5x+5 =x 6 +2x 5 −x 4 +x 3 +x 2 −5x+5 Hence, solve. |
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