# Multiply the monomial by the binomial and find the value of each for $x = -1, y = 0.25$ and $z =0.05$:(i) $15y^2 (2 - 3x)$(ii) $-3x (y^2 + z^2)$(iii) $z^2 (x - y)$(iv) $xz (x^2 + y^2)$

To do:

We have to multiply the monomial by the binomial and find the value of each for $x = -1, y = 0.25$ and $z =0.05$:

Solution:

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

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

If $x=-1, y=0.25$, then

$30 y^{2}-45 x y^{2}=30(0.25)^{2}-45 \times(-1)(0.25)^{2}$

$=30 \times 0.0625+45 \times 0.0625$

$=1.8750+2.8125$

$=4.6875$

$=\frac{46875}{10000}$

$=\frac{75}{16}$

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

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

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

$=3 \times 0.0625+3 \times 0.0025$

$=0.1875+0.0075$

$=0.1950$

$=\frac{1950}{10000}$

$=\frac{39}{200}$

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

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

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

$=-0.0025-0.000625$

$=-0.003125$

$=-\frac{3125}{1000000}$

$=-\frac{5}{1600}$

$=-\frac{1}{320}$

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

$=x^{1+2} z+x y^{2} z$

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

If $x=-1, y=0.25, z=0.05$, then

$x^{3} z+x y^{2} z=(-1)^{3} \times(0.05)+(-1)(0.25)^{2}(0.05)$

$=-1 \times 0.05-1 \times(0.0625) \times(0.05)$

$=-0.05-0.003125$

$=-0.050000-0.003125$

$=-0.053125$

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Updated on: 10-Oct-2022

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