Verify that 3,-2,1 are the zeros of the cubic polynomial P(x)=x³-2x²-5x+6 and verify the relation between its zeros and coefficients.
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Verify that 3,-2,1 are the zeros of the cubic polynomial P(x)=x³-2x²-5x+6 and verify the relation between its zeros and coefficients.
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3
3 −2x
3 −2x 2
3 −2x 2 −5x+6)
Let f(x)=x
Let f(x)=x 3
Let f(x)=x 3 −2x
Let f(x)=x 3 −2x 2
Let f(x)=x 3 −2x 2 −5x+6
Let f(x)=x 3 −2x 2 −5x+63,−2
Therefore,
Therefore,f(3)=(3)
Therefore,f(3)=(3) 3
Therefore,f(3)=(3) 3 −2(3)
Therefore,f(3)=(3) 3 −2(3) 2
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2)
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2)
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1)
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1)
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+d
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+dnow ,
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+dnow ,Consider α=3,β=−2 and y=1
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+dnow ,Consider α=3,β=−2 and y=1α+β+y=3−2+1=2=−b/a
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+dnow ,Consider α=3,β=−2 and y=1α+β+y=3−2+1=2=−b/aαβ+βy+αy=3(−2)+(−2)(1)+1(3)=−5=c/a
Therefore,f(3)=(3) 3 −2(3) 2 −5(−2)+6=27−18−15+6=0f(−2)=(−2) 3 −2(−2) 2 −5(−2)+6−8−8+10+6=0f(1)=(1) 3 −2(1) 2 −5(1)+6=1−2−5+6=0Verify relations :General form of cubic equation :ax 3 +bx 2 +cx+dnow ,Consider α=3,β=−2 and y=1α+β+y=3−2+1=2=−b/aαβ+βy+αy=3(−2)+(−2)(1)+1(3)=−5=c/aand αβy=3(−2)(1)=−6=−d/a