Question

verify that the no. given alongside of the cubic polynomial below r their zeros. Also verify the relationship between the zeros and the coofficients in each case :

1.) p( x ) = 2x³ + x² - 5x +2 ;
1/2,1,-2

2.) p ( x ) = x³ - 4x² + 5x - 2 ;
2,1,1

Ans it asap
Points :- 15

Answer 1

(1)p(x)=2x³+x²-5x+2

Put x=1/2,1,-2 and check whether they are zeroes or not.

p(1/2)=2(1/2)³+(1/2)²-5(1/2)+2
=2(1/8)+1/4-5/2+2
=1/4+1/4-5/2+2
=(1+1-10+8)/4
=0/4=0

p(1)=2(1)³+(1)²-5(1)+2
=2(1)+1-5+2
=2+1-5+2
=0

p(-2)=2(-2)³+(-2)²-5(-2)+2
=2(-8)+4+10+2
=-16+16
=0

Therefore, 1/2,1 and -2 are the zeroes of the given polynomial.

Relation between zeroes and coefficients are:-

Sum of zeroes=1/2+1-2=1/2-1=(1-2)/2=-1/2=-x² coefficient/x³ coefficient

Sum of the product of each two zeroes=(1/2×1)+(1×-2)+(1/2×(-2))=(1/2)+(-2)+(-1)=1/2-2-1=(1-4-2)/2=-5/2=x coefficient/x³ coefficient

Product of all zeroes=(1/2)(1)(-2)= -2/2= - constant/x³ coefficient

(2)p(x)=x³-4x²+5x-2

Put x=2,1,1 and check whether they are zeroes or not.

p(2)=2³-4(2)²+5(2)-2
=8-4(4)+10-2
=8-16+8
=16-16=0

p(1)=1³-4(1)²+5(1)-2
=1-4+5-2
=6-6=0

p(1)=1³-4(1)²+5(1)-2
=1-4+5-2
=6-6=0

Therefore, 2,1,1 are the zeroes of the given polynomial.

Relation between zeroes and coefficients:-

Sum of zeroes=2+1+1=4/1=-(-4)/1= -x² coefficient/x³ coefficient

Sum of product of each two zeroes=(2×1)+(1×1)+(2×1)=2+1+2=5/1
=x coefficient/x³ coefficient

Product of all zeroes=(2)(1)(1)=2/1= -(-2)/1 = - constant/x³ coefficient

Hope it helps

Answer 2

HI there !

p(x)=2x³+x²-5x+2



p(1/2) =2(1/2)³+(1/2)²-5(1/2)+2
=2(1/8)+1/4-5/2+2
=1/4+1/4-5/2+2
=(1+1-10+8)/4
=0/4=0

1/2 is a zero

p(1)=2(1)³+(1)²-5(1)+2
=2 x 1 x +1-5+2
=2+1-5+2
= 0

1 is also a zero

p(-2)=2(-2)³+(-2)²-5(-2)+2
=2(-8)+4+10+2
=-16 + 16
=0
-2 also is a zero of the polynomial

Relation between zeroes and coefficients are:-

Sum of zeroes =-x² coefficient/x³ coefficient
= 1/2+1 -2
=1/2-1
=(1-2)/2
=-1/2

Sum of the product of each two zeroes=x coefficient/x³ coefficient
= (1/2×1)+(1×-2)+(1/2×(-2))
=(1/2)+(-2)+(-1)
=1/2-2-1
=(1-4-2)/2
=-5/2

Product of all zeroes= - constant/x³ coefficient
= (1/2)(1)(-2)
= -2/2

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2)p(x)=x³-4x²+5x-2


p(2)=2³-4(2)²+5(2)-2
=8-4 x 4 x +10-2
=8-16+8
=16-16=0
2 is a zero of the polynomial

p(1)=1³-4(1)²+5(1)-2
=1-4+5-2
=6-6=0
1 is a zero of the polynomial


p(1)=1³-4(1)²+5(1)-2
=1-4+5-2
=6-6=0

1 is a zero of the polynomial



Relation between zeroes and coefficients:-

Sum of zeroes= -x² coefficient/x³ coefficient = 2+1+1=4/1
=-(-4)/1

Sum of product of each two zeroes=x coefficient/x³ coefficient
= (2×1)+(1×1)+(2×1)
=2+1+2
=5/1

Product of all zeroes= - constant/x³ coefficient
= (2)(1)(1)
=2/1
= -(-2)/1

Hope this Helps You !!