Question

Factorise 4z3 + 23z2 - 41x - 42

Answer 1

Answer: (z-2)(z+7)(4z+3)

Step-by-step explanation:

1) We have,

$p(z) = 4z^3+23z^2-41z-42$

By inspection, z = 2 is zero of polynomial p(z) .

So, (z-2) will be a factor of p(z) by Factor theorem.

2) Again,

Making terms in form of above factor.

That is,

$4z^3+23z^2-41z-42\\ \\=4z^3-8z^2+8z^2+23z^2-41z-42\\ \\=4z^2(z-2)+31z^2-62z+62z-41z-42\\ \\=4z^2(z-2)+31z(z-2)+21z-42\\ \\=4z^2(z-2)+31z(z-2)+21(z-2)\\ \\=(z-2)(4z^2+31z+21)\\ \\=(z-2)(4z^2+28z+3z+21)\\ \\=(z-2)(4z(z+7)+3(z+7))\\ \\=(z-2)(z+7)(4z+3)$

Hence, required factorization is

(z-2)(z+7)(4z+3)

Answer 2

Step-by-step explanation:

Answer: (z-2)(z+7)(4z+3)

Step-by-step explanation:

1) We have,

p(z) = 4z^3+23z^2-41z-42p(z)=4z

3

+23z

2

−41z−42

By inspection, z = 2 is zero of polynomial p(z) .

So, (z-2) will be a factor of p(z) by Factor theorem.

2) Again,

Making terms in form of above factor.

That is,

$$\begin{lgathered}4z^3+23z^2-41z-42\\ \\=4z^3-8z^2+8z^2+23z^2-41z-42\\ \\=4z^2(z-2)+31z^2-62z+62z-41z-42\\ \\=4z^2(z-2)+31z(z-2)+21z-42\\ \\=4z^2(z-2)+31z(z-2)+21(z-2)\\ \\=(z-2)(4z^2+31z+21)\\ \\=(z-2)(4z^2+28z+3z+21)\\ \\=(z-2)(4z(z+7)+3(z+7))\\ \\=(z-2)(z+7)(4z+3)\end{lgathered}$$

Hence, required factorization is

(z-2)(z+7)(4z+3)