Question -
Answer -
(i) 2x3+x2-5x+2; -1/2,1, -2
Solution:
Given, p(x) = 2x3+x2-5x+2
And zeroes for p(x) are = 1/2, 1, -2
∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2
= (1/4)+(1/4)-(5/2)+2
= 0
p(1)= 2(1)3+(1)2-5(1)+2
= 0
p(-2)= 2(-2)3+(-2)2-5(-2)+2
= 0
Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.
Now, comparing the given polynomial withgeneral expression, we get;
∴ ax3+bx2+cx+d
= 2x3+x2-5x+2
a=2, b=1, c= -5 and d = 2
As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
Therefore, putting the values of zeroes of thepolynomial,
α+β+γ = ½+1+(-2)
= -1/2
= –b/a
αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2)
= -5/2
= c/a
α β γ = ½×1×(-2)
= -2/2
= -d/a
Hence, the relationship between the zeroes andthe coefficients are satisfied.
(ii) x3-4x2+5x-2 ;2, 1, 1
Solution:
Given, p(x) = x3-4x2+5x-2
And zeroes for p(x) are 2,1,1.
∴ p(2)= 23-4(2)2+5(2)-2
= 0
p(1) = 13-(4×12 )+(5×1)-2
= 0
Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2
Now, comparing the given polynomial withgeneral expression, we get;
∴ ax3+bx2+cx+d = x3-4x2+5x-2
a = 1, b = -4, c = 5 and d = -2
As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;
α + β + γ = –b/a
αβ + βγ + γα = c/a
α β γ = – d/a.
Therefore, putting the values of zeroes of thepolynomial,
α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a
αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a
αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a
Hence, the relationship between the zeroes andthe coefficients are satisfied.