# Multiply the monomial by the binomial and find the value of each for x = −1,

Question:

Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:

(i) 15y2(2 − 3x)

(ii) −3x(y2 + z2)

(iii) z2(x − y)

(iv) xz(x2 + y2)

Solution:

(i) To find the product, we will use distributive law as follows:

$15 y^{2}(2-3 x)$

$=15 y^{2} \times 2-15 y^{2} \times 3 x$

$=30 y^{2}-45 x y^{2}$

Substituting $x=-1$ and $y=0.25$ in the result, we get:

$30 y^{2}-45 x y^{2}$

$=30(0.25)^{2}-45(-1)(0.25)^{2}$

$=30 \times 0.0625-\{45 \times(-1) \times 0.0625\}$

$=30 \times 0.0625-\{45 \times(-1) \times 0.0625\}$

$=1.875-(-2.8125)$

$=1.875+2.8125$

$=4.6875$

(ii) To find the product, we will use distributive law as follows:

$-3 x\left(y^{2}+z^{2}\right)$

$=-3 x \times y^{2}+(-3 x) \times z^{2}$

$=-3 x y^{2}-3 x z^{2}$

Substituting $x=-1, y=0.25$ and $z=0.05$ in the result, we get:

$-3 x y^{2}-3 x z^{2}$

$=-3(-1)(0.25)^{2}-3(-1)(0.05)^{2}$

$=-3(-1)(0.0625)-3(-1)(0.0025)$

$=01875+0.0075$

$=0.195$

(iii) To find the product, we will use distributive law as follows:

$z^{2}(x-y)$

$=z^{2} \times x-z^{2} \times y$

$=x z^{2}-y z^{2}$

Substituting $x=-1, y=0.25$ and $z=0.05$ in the result, we get:

$x z^{2}-y z^{2}$

$=(-1)(0.05)^{2}-(0.25)(0.05)^{2}$

$=(-1)(0.0025)-(0.25)(0.0025)$

$=-0.0025-0.000625$

$=-0.003125$

(iv) To find the product, we will use distributive law as follows:

$x z\left(x^{2}+y^{2}\right)$

$=x z \times x^{2}+x z \times y^{2}$

$=x^{3} z+x y^{2} z$

Substituting $x=-1, y=0.25$ and $z=0.05$ in the result, we get:

$x^{3} z+x y^{2} z$

$=(-1)^{3}(0.05)+(-1)(0.25)^{2}(0.05)$

$=(-1)(0.05)+(-1)(0.0625)(0.05)$

$=-0.05-0.003125$

$=-0.053125$