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CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 5 - MCQExams.com
CBSE
Class 7 Maths
Algebraic Expressions
Quiz 5
The sum of three algebric expressions is $$\displaystyle x^{2}+y^{2}+z^{2}$$. If two of them are $$\displaystyle 4x^{2}-5y^{2}+3z^{2}$$ and $$\displaystyle -3x^{2}+4y^{2}-2z^{2}$$ then the third expression is
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$$\displaystyle 2x^{2}+2z^{2}$$
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$$\displaystyle 2y^{2}$$
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$$\displaystyle 2x^{2}+2y^{2}-z^{2}$$
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$$\displaystyle 2y^{2}+2z^{2}$$
Explanation
Third exp = $$\displaystyle x^{2}+y^{2}+z^{2}-\left \{ \left ( 4x^{2}-5y^{2}+3z^{2} \right )+\left ( -3x^{2}+4y^{2}-2z^{2} \right ) \right \}$$
= $$x^2+y^2+z^2 - ( x^2 - y^2 + z^2) $$
= $$ x^2 + y^2 + z^2 - x^2 + y^2 - z^2$$
= $$2y^2$$
If $$\displaystyle P=3x-4y-8z,\:Q=-10y+7x+11z$$ and $$\displaystyle R=19z-6y+4x$$,
then $$P - Q + R$$ is equal to
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$$\displaystyle 13x-20y+16z$$
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$$0$$
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$$x + y + z$$
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$$\displaystyle 2x-4y+3z$$
Explanation
Given, $$P=3x-4y-8z, Q=-10y+7x+11z, R=19z-6y+4x$$
We need to find value of $$P - Q + R $$
$$= (3x - 4y - 8z) - (-10y + 7x + 11z) + (19z - 6y + 4x)$$
$$ = 3x - 4y - 8z + 10y - 7x - 11z + 19z - 6y + 4x$$
$$= (3x-7x+4x)+(-4y+10y-6y)+(-8z-11z+19z)$$
$$=0+0+0=0$$
$$\displaystyle 5a-\left[ 3b-\left\{ a-3\left( 2a-b \right) \right\} \right] $$ is equal to
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$$\displaystyle -2b$$
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$$\displaystyle -a+b$$
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$$0$$
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$$\displaystyle -a$$
Explanation
Given, $$\displaystyle 5a-\left[ 3b-\left\{ a-3\left( 2a-b \right) \right\} \right] $$
$$\displaystyle =5a-\left[ 3b-\left\{ a-6a+3b \right\} \right] $$
$$\displaystyle =5a-\left[ 3b-\left\{ -5a+3b \right\} \right] $$
$$\displaystyle =5a-\left[ 3b+5a-3b \right] $$
$$\displaystyle =5a-5a=0$$
Hence given expression vanishes.
Simplify: $$\displaystyle x^{2}y^{3}-1.5x^{2}y^{3}+1.4x^{2}y^{3}$$
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$$\displaystyle 0.9x^{2}y^{3}$$
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$$\displaystyle -0.9x^{2}y^{3}$$
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$$0.9$$
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$$-0.9$$
Explanation
$$\displaystyle x^{2}y^{3}-1.5x^{2}y^{3}+1.4x^{2}y^{3}$$
=$$\displaystyle x^{2}y^{3}(1-1.5+1.4)$$
$$\displaystyle =0.9x^{2}y^{3}$$
What must be added to $$x^2\,+\, 5x\,-\,6$$ to get $$x^3\, -\,x^2\,+\, 3x\, -\, 2$$?
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$$x^3\,-\,2x^2\,-\,2x\,-\,4$$
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$$x^3\,+\,2x^2\,-\,2x\,+\,4$$
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$$x^3\,-\,2x^2\,-\,2x\,+\,4$$
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None of these
Explanation
Let the polynomial to be added be $$p(x)$$
$$\therefore x^2 + 5x -6 +p(x) = x^3-x^2 +3x-2$$
$$\therefore p(x) = x^3 -x^2-x^2+3x-5x-2+6$$
$$\therefore p(x) = x^3 -2x^2-2x+4$$
What must be subtracted from $$x^4\, +\, 2x^2\,-\,3x\, +\, 7$$ to get $$x^3\,+\, x^2\,+\, x\, -\,1$$?
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$$x^4\, -\,x^3\,+ x^2\, -\, 4x\, +\, 8$$
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$$x^3\,+\, x^2\, -\, 4x\, +\, 8$$
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$$x^4\, -\,x^3\, +\, x^2\,+\, 4x\, -\,8$$
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$$x^4\, -\,x^3\, -\,x^2\,+\, 4x\, -\,8$$
Explanation
Let the polynomial to be subtracted be $$p(x)$$
$$\therefore x^4 +2x^2 - 3x +7 -p(x) = x^3 + x^2 + x -1$$
$$\therefore x^4-x^3+2x^2-x^2-3x-x+7+1 = p(x)$$
$$\therefore p(x) = x^4 - x^3 + x^2 - 4x + 8$$
What must be subtracted from
$$x^3 \,-\, 3x^2\, + 5x\, -\, 1$$ to get $$2x^3\, +\, x^2\, - \,4x\, +\, 2$$?
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$$-x^3\, +\, 4x^2\, -\, 9x\, +\, 3$$
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$$x^3\, +\, 4x^2\, -\, 9x\, +\, 3$$
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$$x^3\, -\, 4x^2\, +\, 9x\, -\, 3$$
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$$-x^3\, -\, 4x^2\, +\, 9x\, -\, 3$$
Explanation
Let the polynomial to be subtracted be $$p(x)$$.
$$\therefore x^3 - 3x^2 + 5x -1 - p(x) = 2x^3 + x^2 - 4x + 2$$
$$x^3-2x^3 - 3x^2 - x^2 +5x + 4x-1-2 = p(x)$$
$$\therefore p(x) = -x^3 - 4x^2 + 9x -3$$
What must be added to
$$x^3\, +\, 3x\, -\, 8$$ to get $$3x^3 \,+ \,x^2\, +\, 6$$?
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$$2x^3\, +\, x^2\, - \,3x\, +\, 14$$
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$$2x^2\, + \, x^2\, +\, 14$$
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$$2x^3\, +\, x^2\, - \,6x\, -\,14$$
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None of these
Explanation
Let the polynomial to be added be $$p$$.
$$\therefore x^3 + 3x - 8 + p = 3x^3 + x^2 + 6$$
$$ \therefore p = 3x^3 + x^2 + 6 - x^3 - 3x + 8$$
$$\therefore p = 2x^3 + x^2- 3x + 14$$
Hence, Option A is correct.
What must be added to the sum of $$\displaystyle 2a^{2}-3a+7,-5a^{2}-2a-11$$ and $$\displaystyle 3a^{2}+5a-8$$ to get $$0$$?
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$$-12$$
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$$12$$
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$$\displaystyle a^{2}+a$$
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$$\displaystyle a-1$$
Explanation
Let '$$x$$' be added to these polynomial to get $$0$$.
$$\Rightarrow (2a^2-3a+7)+(-5a^2-2a-11)+(3a^2+5a-8)+x=0$$
$$\Rightarrow (2a^2-5a^2+3a^2)+(-3a-2a+5a)+(7-11-8)+x=0$$
$$\Rightarrow 0+0+(-12)+x=0$$
$$\Rightarrow x=12$$
Option B is correct.
Which of the following expressions are exactly equal in value?
$${(3x-y)}^{2}-({5x}^{2}-2xy)$$
$${(2x-y)}^{2}$$
$${(2x+y)}^{2}-2xy$$
$${(2x+3y)}^{2}-8y(2x+y)$$
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$$1$$ and $$2$$ only
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$$1,2$$ and $$3$$ only
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$$2$$ and $$4$$ only
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$$1,2$$ and $$4$$ only
Explanation
From $$1^{st}$$ expression
$${(3x-y)}^{2}-({5x}^{2}-2xy)$$
$$=9{x}^{2}-6xy+{y}^{2}-5{x}^{2}+2xy$$
$$=4{x}^{2}-4xy+{y}^{2}$$
From $$2^{nd}$$ expression
$${(2x-y)}^{2} =4{x}^{2}-4xy+{y}^{2}$$
From $$3^{rd}$$ expression
$${ \left( 2x+y \right) }^{ 2 }-2xy=4{ x }^{ 2 }+4xy+{ y }^{ 2 }-2xy$$
$$=4{ x }^{ 2 }+2xy+{ y }^{ 2 }\quad $$
From $$4^{th}$$ expression
$${(2x+3y)}^{2}-8y(2x+y)=4{ x }^{ 2 }+12xy+{ 9y }^{ 2 }-16xy-8{ y }^{ 2 }$$
$$=4{ x }^{ 2 }-4xy+{ y }^{ 2 }$$
Hence the value of $$1,2$$ and $$4$$ expression will be same
Subtract:
$$-6p+ q+3r+ 8$$ from $$p -2q - 5r -8$$
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$$7p-3q+8r-16$$
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$$7p-3q-8r-16$$
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$$p-3q-8r+16$$
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$$p-3q-8r-16$$
Explanation
$$\begin{aligned}{}p - 2q - 5r - 8 - ( - 6p + q + 3r + 8) &= p - 2q - 5r - 8 + 6p - q - 3r - 8\\ &= p + 6p - 2q - q - 5r - 3r - 8 - 8\\& = 7p - 3q - 8r - 16\end{aligned}$$
Hence, option $$B$$ is correct.
What should be added to the product of $$({x}^{2}+xy-{y}^{2})$$ and $$({x}^{2}-xy+{y}^{2})$$ to get $${x}^{2}{y}^{2}$$?
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$$2{x}^{2}{y}^{2}-{x}^{4}-2x{y}^{3}+{y}^{4}$$
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$$2xy-x^{3}+2xy+y^{4}$$
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$$2x^2y^2+x^4+2xy^3-y^4$$
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$$2x^{2}y^{2}+x^{4}+2xy^{3}+y^{4}$$
Explanation
The given expressions are $$({x}^{2}+xy-{y}^{2})$$ and $$({x}^{2}-xy+{y}^{2})$$
There product $$= ({x}^{2}+xy-{y}^{2})({x}^{2}-xy+{y}^{2})$$
$$= {x}^{2}({x}^{2}-xy+{y}^{2})+xy({x}^{2}-xy+{y}^{2})-{y}^{2}({x}^{2}-xy+{y}^{2})$$
$$= \left ( x^4-x^3y+x^2y^2 \right )+\left(x^3y-x^2y^2+xy^3 \right )+\left (-x^2y^2+xy^3-y^4 \right )$$
$$= {x}^{4}+2x{y}^{3}-{x}^{2}{y}^{2}-{y}^{4}$$
let we added $$a$$ then
$$ a={x}^{2}{y}^{2}-({x}^{4}+2x{y}^{3}-{x}^{2}{y}^{2}-{y}^{4})$$
$$\Rightarrow {x}^{2}{y}^{2}-{x}^{4}-2x{y}^{3}+{x}^{2}{y}^{2}+{y}^{4}$$
$$\Rightarrow 2{x}^{2}{y}^{2}-{x}^{4}-2x{y}^{3}+{y}^{4}$$
Subtract $$(4a -7ab + 3b +12)$$ from $$(12a -9ab+5b -3) .$$
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$$8a - 2ab + 2b -15$$
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$$a - ab + 2b -25$$
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$$a - 2ab + 2b -5$$
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$$8a - ab + 2b -10$$
Explanation
To s
ubtract $$(4a -7ab + 3b +12)$$ from $$(12a -9ab+5b -3) .$$
Rearranging the terms of the given expressions, changing the sign of each term of the expression
to be subtracted and adding the two expressions, we get.
$$12a - 9ab + 5b -3$$
$$4a - 7ab + 3b +12$$
$$ -$$ $$ +$$ $$- $$ $$ -$$
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$$8a - 2ab + 2b -15$$
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Hence, $$Op-A$$ is correct .
Subtract:
$$5y^4-3y^3+2y^2+y-1$$ from $$4y^4- 2y^3 -6y^2-y + 5$$
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$$-y^4 + y^3-8y^2-2y + 6$$
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$$y^4 + y^3-8y^2-2y + 4$$
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$$-y^4 + y^3-8y^2-2y - 4$$
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$$y^4 + y^3-8y^2+2y + 4$$
Explanation
After subtracting $$5y^4-3y^3+2y^2+y-1$$ from, $$4y^4 -2y^3 -6y^2-y + 5$$ we get,
$$4y^4- 2y^3- 6y^2-y + 5-(5y^4-3y^3+2y^2+y-1)$$
$$=4y^4- 2y^3- 6y^2-y + 5-5y^4+3y^3-2y^2-y+1$$
$$=-y^4 + y^3-8y^2-2y + 6$$
$$(5x^{2} + 6x - 3) + (2x^{2} - 7x - 9)$$
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$$7x^{2} - x - 12$$
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$$7x^{2} - 2x - 12$$
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$$7x^{2} - 3x - 12$$
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None of the above
Explanation
$$5x^{2} + 6x - 3$$
$$+\underline { 2x^{2} - 7x - 9}$$
$$7x^{2} - x - 12$$
The total terms in $$-x^3+4x^2+7x-2$$ is
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2
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3
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4
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5
Explanation
We have a polynomial given by $$-x^3 + 4x^2 + 7x - 2$$
The total number of terms is given by 4, since we have 4 terms with either an addition or subtraction operated between them.
Those 4 terms are $$-x^3, 4x^2, 7x, -2$$
What is to be added to $$2a^3+4ab^2-5b^3-6a^2b$$ to get $$3a^3-4a^2b+11ab^2+b^3$$?
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$$a^3 + 2a^2b+ 7ab^2-6b^3$$
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$$a^3 -2a^2b-7ab^2+ 6b^3$$
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$$a^3+2a^2b+7ab^2+6b^3$$
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$$a^3+2a^2b-7ab^2+6b^3$$
Explanation
$$3a^3-4a^2b+11ab^2+b^3$$
$$\underline {-2a^3+6a^2b-4ab^2+5b^3}$$
$$a^3+2a^2b+7ab^2+6b^3$$
What must be subtracted from $$3a^2-6ab-3b^2-1 $$ to get $$4a^2-7ab -4b^2 + 1?$$
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$$-a^2 + ab + b^3-2$$
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$$-a^2 + ab + b^2-2$$
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$$a^2 + ab + b^2-2$$
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$$a^2 + ab + b^3-2$$
Explanation
Let $$X$$ be subtracted from $$3a^2-6ab-3b^2$$
Then,
$$3a^2-6ab-3b^2-1 -X= 4a^2-7ab -4b^2 + 1$$
$$x=3a^2-6ab-3b^2-1-(4a^2-7ab -4b^2 + 1)$$
$$x=3a^2-6ab-3b^2-1-4a^2+7ab +4b^2 - 1$$
$$x=-a^2+ab+b^2-2$$
How many terms are there in the expression $$5x^3+3xy+y^2$$ ?
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$$1$$
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$$2$$
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$$3$$
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$$5$$
Explanation
There are 3 terms. The terms are $$5x^3, 3xy, y^2$$
If we add two binomials $$4x^2+5y^2$$ and $$7x^2+y^2$$, what will be the result?
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monomial
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binomial
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trinomial
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cannot be determined
Explanation
$$(4x^2+5y^2)+(7x^2+y^2)=11x^2+6y^2$$ is a binomial.
What is the measure of the third side of a triangle given that its two sides are $$a^{2} - 2a + 1$$ and $$3a^{2} - 5a + 3$$ and has a perimeter $$6a^{2} - 4a + 9$$?
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$$2a^{2} - 3a - 5$$
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$$2a^{2} + 3a - 5$$
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$$2a^{2} + 3a + 5$$
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$$2a^{2} - 3a + 5$$
Explanation
Let the third side of the triangle be $$x.$$
The perimeter of a triangle is equal to the sum of its sides so,
$$\begin{aligned}{}x + \left( {{a^2} - 2a + 1} \right) + \left( {3{a^2} - 5a + 3} \right) &= 6{a^2} - 4a + 9\\x + \left( {{a^2} + 3{a^2}} \right) + \left( { - 2a - 5a} \right) + \left( {1 + 3} \right)& = 6{a^2} - 4a + 9\\x + 4{a^2} - 7a + 4 &= 6{a^2} - 4a + 9\\x &= \left( {6{a^2} - 4{a^2}} \right) + \left( { - 4a + 7a} \right) + \left( {9 - 4} \right)\\& = 2{a^2} + 3a + 5\end{aligned}$$
Hence, option $$C$$ is correct.
$$3x^{2} - 5x + 2$$
$$5x^{2} - 2x - 6$$
Which of the following is the sum of the two polynomials shown above?
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$$8x^{2} - 7x - 4$$
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$$8x^{2} + 7x - 4$$
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$$8x^{4} - 7x^{2} - 4$$
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$$8x^{4} + 7x^{2} - 4$$
Explanation
Given polynomials are $$3x^2-5x+2$$ and $$5x^2-2x-6$$
The sum of the given polynomials is given by:
$$(3x^2-5x+2)+(5x^2-2x-6)$$
$$=(3x^2+5x^2)+(-5x-2x)+(2-6)$$
$$=(3+5)x^2 -(5+2)x -4$$
$$=8x^2-7x-4$$
If $$P=3x-4y-8z$$, $$Q=-10y+7x+11z$$ and $$R=19z-6y+4x$$, then $$P-Q+R$$ is equal to
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$$13x-20y+16z$$
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$$0$$
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$$x+y+z$$
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$$2x-4y+3z$$
Explanation
$$P=3x-4y-8z$$
$$Q=-10y+7x+11z$$
$$R=19z-6y+4x$$
$$\therefore P-Q+R=(3x-4y-8z)-(-10y+7x+11z)+(19z-6y+4x)$$
$$=3x-4y-8z+10y-7x-11z+19z-6y+4x$$
$$=3x-7x+4x-4y+10y-6y-8z-11z+19z=7x-7x-10y+10y-19z+19z=0$$
Simplify the polynomial and write it in standard form:
$$-3({x}^{3}-{x}^{2}-2x-5)-(4{x}^{3}-7x-1)$$
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$$-7{x}^{3}+3{x}^{2}+13x+16$$
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$$7{x}^{3}+2{x}^{2}+11x+16$$
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$$-6{x}^{3}+3{x}^{2}+12x+14$$
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$$-4{x}^{3}+3{x}^{2}+11x+15$$
Explanation
Solve the polynomial as follows:
$$-3(x^3-x^2-2x-5)-(4x^3-7x-1)$$
$$=-3x^3+3x^2+6x+15-4x^3+7x+1$$
$$=-7x^3+3x^2+13x+16$$
The sum of three expressions is $${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$$. If two of them are $$4{ x }^{ 2 }-5{ y }^{ 2 }+3{ z }^{ 2 }$$ and $$-3{ x }^{ 2 }+4{ y }^{ 2 }+2{ z }^{ 2 }$$, the third expression is
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$$2{ x }^{ 2 }+2{ z }^{ 2 }$$
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$$2{ y }^{ 2 }$$
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$$2{ x }^{ 2 }+2{ y }^{ 2 }-{ z }^{ 2 }$$
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$$2{ y }^{ 2 }-4{ z }^{ 2 }$$
Explanation
Add two expressions
$$\Rightarrow \left ( 4x^{2}-5y^{2}+3z^{2} \right )+(-3x^{2}+4y^{2}+2z^{2})$$
$$=\left ( 4x^{2}-5y^{2}+3z^{2}-3x^{2}+4y^{2}+2z^{2} \right )$$
$$=4x^{2}-3x^{2}-5y^{2}+4y^{2}+3z^{2}+2z^{2}$$
$$=x^{2}-y^{2}+5z^{2}$$
But
$$x^{2}-y^{2}+5z^{2}+P(x)=x^{2}+y^{2}+z^{2}$$, where $$P(x)$$ is the third expression
$$\therefore P(x)=\left ( x^{2}+y^{2}+z^{2} \right )-\left ( x^{2}-y^{2}+5z^{2} \right )$$
$$=x^{2}+y^{2}+z^{2}-x^{2}+y^{2}-5z^{2}=2y^{2}-4z^{2}$$
Add $$(12x^2+4x+5y)$$ and $$(3x^2-2x+3y)$$.
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$$2x^2-2x+8y$$
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$$2x^2+15x+8y$$
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$$15x^2-2x+8y$$
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$$15x^2+2x+8y$$
Explanation
Adding $$ (12x^2+4x+5y)+(3x^2-2x+3y)$$
$$\Rightarrow 12x^2+4x+5y+3x^2-2x+3y$$
$$\Rightarrow 15x^2+2x+8y$$
$$x+y-\left( z-x-\left[ y+z-\left( x+y-\left\{ z+x-\left( y+z+x \right) \right\} \right) \right] \right) $$ is equal to
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$$3x$$
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$$2y$$
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$$x$$
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$$0$$
Explanation
$$x+y-\left (z-x-\left [ y+z-(x+y-\left \{ z+x-(y+z+x) \right \}) \right ] \right )$$
$$=x+y-\left ( z-x-\left [ y+z-(x+y-\left \{ z+x-y-z-x \right \}) \right ] \right )$$
$$=x+y-\left (z-x-\left [ y+z-(x+y-z-x+y+z+x) \right ] \right )$$
$$=x+y-\left ( z-x-\left [ y+z-x-y+z+x-y-z-x \right ] \right )$$
$$=x+y-\left (z-x-y-z+x+y-z-x+y+z+x \right )$$
$$=x+y-z+x+y+z-x-y+z+x-y-z-x$$
$$=x$$
Simplify: $$\left( 5{ x }^{ 2 }-3x+2 \right) -\left( { 3x }^{ 2 }+5x-1 \right) +\left( 6{ x }^{ 2 }-2 \right) $$
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$$5{ x }^{ 2 }-3x+5$$
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$$6{ x }^{ 2 }-4x+1$$
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$$8{ x }^{ 2 }-8x+1$$
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$$8{ x }^{ 2 }-9x+5$$
Explanation
The value of
$$(5{x}^{2}-3x+2)-(3{x}^{2}+5x-1)+(6{x}^{2}-2)$$
$$ = (5{x}^{2}-3{x}^{2}+6{x}^{2})+(-3x-5x)+(2+1-2) $$
$$= 8{x}^{2}-8x+1$$
What is the simplified result of following the steps in order?
$$(1)$$ add $$5y$$ to $$2x$$
$$(2)$$ multiply the sum by $$3$$
$$(3)$$ subtract x$$+$$y from the product
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$$5x+14y$$
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$$5x+16y$$
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$$5x+5y$$
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$$6x+4y$$
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$$3x+12y$$
$$\left( 4p{ x }^{ 2 }+5{ q }^{ 2 }y-9rz \right) -\left( -3{ q }^{ 2 }y+7p{ x }^{ 2 }-rz \right) =$$
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$$11p{ x }^{ 2 }+9{ q }^{ 2 }y-10rz$$
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$$-3p{ x }^{ 2 }+8q^{2}y-8rz$$
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$$11p{ x }^{ 2 }+8{ q }^{ 2 }y-8rz$$
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$$-3p{ x }^{ 2 }-8{ q }^{ 2 }y-10rz$$
Explanation
$$(4px^{2}+5q^{2}y-9rz)-(-3q^{2}y+7px^{2}-rz)$$
$$=4px^{2}+5q^{2}y-9rz+3q^{2}y-7px^{2}+rz$$
$$=4px^{2}-7px^{2}+5q^{2}y+3q^{2}y-9rz+rz$$
$$=-3px^{2}+8q^{2}y-8rz$$
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