Question

verify that the number given alongside of the cubic polynomial below are their zerores x3-4x2+5x-2;2,1,1 ​

Answer 1

Step-by-step explanation:

Let p(x) = x3 – 4x2 + 5x – 2

Then, p(2) = (2)3 – 4(2)

2 + 5(2) – 2 = 8 – 16 + 10 – 2

0 p(1) = (1)3 – 4(1)2 + 5

(1) – 2 = 1 – 4 + 5 – 2

= 0

Answer 2

Step-by-step explanation:

$\huge\bold\green{Solution}$

$1) x³−4 x²+5x−2 ; 2, 1, 1$

$p(x)=x³−4 x²+5x−2 \: \: \: (1)$

• Zero for this polynomial are 2,1,1

$Substitute \\  x=2  \: in \: equation (1)$

${p(2) = {2}^{3} - 4 \times {2}^{2} + 5 \times 2 - 2}$

=8−16+10−2=0

$Substitute \\ \:  x=1  \: in \: equation (1)$

$p(1) = {x}^{3} - 4 {x}^{2} + 5x - 2$

$= {1}^{2} - 4 {(1)}^{2} + 5(1) - 2$

=1−4+5−2=0

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

Comparing the given polynomial with

$a {x}^{2} + b {x}^{2} + cx + d \\ we \: obtain \: \\ a = 1 \: \: b = - 4 \\ c = 5 \: \: d = - 2$

$Let \: us \: assume α=2,  \\ β=1, γ=1$

Sum of the roots = α+β+γ=2+1+1=

$= 4 - \frac{ - 4}{1} \frac{ - b}{a}$

Multiplication of two zeroes taking two at a time=αβ+βγ+αγ=(2)(1)+(1)(1)+(2)(1)

$= 5 = \frac{5}{1} = \frac{c}{a}$

Product of the roots = αβγ=2×1×1=

$2 = - \frac{ - 2}{1} = \frac{d}{a}$

Therefore, the relationship between the zeroes and coefficient are verified.

$\huge\boxed{\dag\sf\red{Thanks}\dag}$