Free Algebraic Expressions and Identities 02 Practice Test - 8th Grade 

Let the expression to be subtracted be $y$.

Hence,
$(x^3-3x^2+5x-1)-y=2x^3+x^2-4x+2$

On rearranging the terms, 
$y=(x^3-3x^2+5x-1)-(2x^3+x^2-4x+2)$

$y=x^3-3x^2+5x-1-2x^3-x^2+4x-2$

$y=x^3-2x^3-3x^2-x^2+5x+4x-1-2$

$\Rightarrow y=-x^3-4x^2+9x-3$

So we have to subtarct $(-x^3-4x^2+9x-3)$ to get the required value.

$\text{To find :}\text{Sum of}(11x^2-8x+4)\text{and}(6x^2+7x-10)$

On adding the like terms together, we get,

$11x^2-8x+4$
$\underset{–}{+6x^2+7x-10}$
$17x^2-x-6$

 

Using the Identity $(x+a)(x-a)=x^2-a^2$

We Put x = 36 and a = 35

(36 + 35) (36 - 35) = (71) (1) = 71

Using the identity,
${(a+b)}^2=a^2+b^2+2ab,$

we get,
${(xy+yz)}^2=x^2{y}^2+y^2{z}^2+2xz{y}^2...\left(1\right)$

Using the identity,
${(a-b)}^2=a^2+b^2-2ab$

we get,
${(xy-yz)}^2=x^2{y}^2+y^2{z}^2-2xz{y}^2...\left(2\right)$

Subtracting (2) from (1), we get

$(x^2{y}^2+y^2{z}^2+2xz{y}^2)-(x^2{y}^2+y^2{z}^2-2xz{y}^2)$
$=2xz{y}^2+2xz{y}^2$
$=4x{y}^{2}z$ 

${(xy+yz)}^2$ is in the form of ${(a+b)}^2$ where $a=xy$ and $b=yz.$
 Using ${(a+b)}^2=a^2+b^2+2ab,$
 ${(xy+yz)}^2=x^2{y}^2+y^2{z}^2+{2xzy}^2.$

Therefore,
${(xy+yz)}^2-2x^2{y}^{2}z=x^2{y}^2+y^2{z}^2+2xz{y}^2-2x^2{y}^{2}z$

Substituting the values of $x$,$y$ and $z$ in the above expression, we get
$x^2{y}^2+y^2{z}^2+2xz{y}^2-2x^2{y}^{2}z$
${(-1)}^2{\left(1\right)}^2+{\left(1\right)}^2{\left(2\right)}^2+2(-1)\left(2\right){\left(1\right)}^2-2{\left(1\right)}^2{\left(1\right)}^2\left(2\right)$
1 + 4 - 4 - 4 = -3

An algebraic expression consists of variables, constants and operators (+, -, / ,x). An algebraic expression is the sum of algebraic terms.

When an algebraic expression is equated to another algebraic expression or zero, it becomes an algebraic equation.

• $\frac{x}{2}-3$ has the variable $x$, constant term $3$ and the operator ${}^{\prime }{-}^{\prime }$.
• $4x+7$ has the variable $x$, constant term $7$ and the operator ${}^{\prime }{+}^{\prime }$.
• $3$ can be written as $3x^0+0$ which has a variable term, constant term and the operator.

Hence, $\frac{x}{2}-3,4x+7$ and $3$ are algebraic expressions where as $x=10$ is an algebraic equation.  

A polynomial which involves two terms is called a binomial. For example, 3y + 9, 4a - 10 etc.

$(a+b+c)(a+b-c)=$
$(a^2+ab-ac+ab+b^2-bc+ac+bc-c^2)$
$=(a^2+2ab+b^2-c^2)$

On multiplying $x$ with $x$, its power is raised to 2 i.e.,  $x\times x=x^2$
On multiplying $z$ with $z$ its power is also raised to 2 .
i.e.,
$z\times z=z^2$
Multiply all the coefficients to get $5\times (-4)\times 3=-60$

Therefore, the answer is $-60x^{2}y{z}^2$.