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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
$x^{3} z+x y^{2} z=(-1)^{3} \times(0.05)+(-1)(0.25)^{2}(0.05)$
$=-1 \times 0.05-1 \times(0.0625) \times(0.05)$
$=-0.05-0.003125$
$=-0.050000-0.003125$
$=-0.053125$