Polynomial equation with real coefficients that has roots

Cheyanne Leigh
2021-01-19
Answered

Polynomial equation with real coefficients that has roots

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broliY

Answered 2021-01-20
Author has **97** answers

Given:) $-1,4-2i$

Used Formula:

$(a+b)(a-b)={a}^{2}-{b}^{2}$

Calculation:

If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.

Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.

Hence, below mentioned equation is written as.

$[x-(-1)][x-(4-2i)][x-(4+2i)]=0$

$(x+1)(x-4+2i)(x-4-2i)=0$

Further use arithmetic rule.

$(a+b)(a-b)={a}^{2}-{b}^{2}$

Here,$a=x-4,b=2i$

Now, the polynomial equation is,

$(x+1)[(x-4{)}^{2}-(2i{)}^{2}]=0$

Use arithmetic rule.

$(a-b{)}^{2}={a}^{2}-2ab+{b}^{2}$

And${i}^{2}=-1$

Now, the polynomial equation is,

$(x+1)({x}^{2}-8x+16+4)=0$

$(x+1)({x}^{2}-8x+20)=0$

$({x}^{3}+{x}^{2}-8{x}^{2}-8x+20x+20)=0$

${x}^{3}-7{x}^{2}+12x+20=0$

Hence, the polynomial equation of giveb roots$-1,4-2iis{x}^{3}-7{x}^{2}+12x+20=0$

Used Formula:

Calculation:

If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4 — 2i must have another root as 4 + 2i.

Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial.

Hence, below mentioned equation is written as.

Further use arithmetic rule.

Here,

Now, the polynomial equation is,

Use arithmetic rule.

And

Now, the polynomial equation is,

Hence, the polynomial equation of giveb roots

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