Question

factorise 4z3 - 20z2 + 7z + 35​

Answer 1

$\huge \mathfrak \red{Answer} \:$

• 4z3 - 20z2 + 7z + 75
• 12z - 40z + 7z + 75
• 12z - 40z + 82z
• 12z - (-122z)
• 134z

Answer 2

$f(z) = 4 {z}^{3} - 20 {z}^{2} + 7z + 35 = 0$

The above cubic equation has no rational zeroes.

So, we are going to use the cubic formula to solve for the roots.

The discriminant of the above cubic equation is:

$Δ = - 27( {4}^{2} )(35 ^{2}) + 18(4)( - 20)7(35) - 4(4) {7}^{3} - 4( { - 20}^{3} )35 + (20 ^{2}.7^{2} ) \\ Δ = 252112 > 0$

As, Δ > 0

So, the roots are all real and distinct.

The three real roots are given by:

$z_{1,2,3} = \frac{10 + \omega_{k} \sqrt[3]{ - 260 + 3i \sqrt{47271} } + \omega_{k} ^{2} .\sqrt[3]{ - 260 - 3i \sqrt{47271} } }{6}$

where, i = √(–1)

On simplifying we get :

$z_{1,2,3} = \frac{5}{3} + \sqrt{79} ( \frac{ \omega_{k}.cis( \frac{\pi}{3} - \frac{1}{3} \tan ^{ - 1} ( \frac{3 \sqrt{47271} }{260} )) + \omega _k ^{2} .cis( - \frac{\pi}{3} + \frac{1}{3} \tan ^{ - 1} ( \frac{3 \sqrt{47271} }{260} )) }{6} )$

where, ωₖ denotes the three cube roots of unity.

and cis(θ)≡cos(θ)+i.sin(θ)

We can now get the linear factors easily which we can mulitply again to get the cubic f(z).

i.e., f(z) = 4(z–z₁)(z−z₂)(z−z₃)

After approximation:

f(z) ≈ 4(z − 4.025)(z − 4.04)(z + 1.065)