Question

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Answer 1

Answer:

4.x

x 3

x 3 −10x

x 3 −10x −53x−42

x 3 −10x −53x−42let p(x)=x

x 3 −10x −53x−42let p(x)=x −10x

x 3 −10x −53x−42let p(x)=x −10x −53x−42

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomial

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x 2

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x 2 −14x+3x−42

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x 2 −14x+3x−42 =x(x−14)+3(x−14)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x 2 −14x+3x−42 =x(x−14)+3(x−14) =(x+3)(x−14)

x 3 −10x −53x−42let p(x)=x −10x −53x−42let x=−1p(−1)=(−1) −10(−1) −53(−1)−42 =−1−10+53−42 =0Hence (x+1) is factor of polynomialp(x)=x 3 +x 2 −11x 2 −42x−11x−42⇒x 2 (x+1)−11x(x+1)−42(x+1)⇒(x+1)(x 2 −11x−42)x 2 −11x−42=x 2 −14x+3x−42 =x(x−14)+3(x−14) =(x+3)(x−14)∴p(x)=(x+1)(x+3)(x−14)

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Answer 2

Answer:

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Explanation:

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