Polynomials and Rational Expressions Questions for Tests and Worksheets

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Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

Multiply the monomials.

$3x^7 * 2x^4$

$6x^11$
$5x^11$
$6x^28$
$5x^28$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

Multiply the monomial with the binomial.

$(-3y^3) (-2y^2 + 9y)$

$6y^5 - 27y^4$
$-6y^5 - 27y^4$
$-5y^5 + 6y^4$
$-6y^6 -27y^3$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

$(-4m-n)-(2m+3n)$

$-2m-4n$
$-6m-4n$
$-6m+2n$
$-8m-2n$

Grade 11 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

What is the product

$(x + 5)(x^3 - 2x -3)?$

$x^4 + 5x^3 - 2x^2 - 7x -15$
$x^4 + 5x^3 - 2x^2 - 13x - 15$
$x^4 + 5x^3 - 2x^2 - 10x - 15$
$x^4 + 5x^3 - 2x^2 - 3x - 15$

Grade 9 Polynomials and Rational Expressions

State the degree of the polynomial.

$5x^4+9x^3+x^2-6x+2$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

Multiply.

$3x(4x+7)$

$12x + 21$
$3x^2 + 7x$
$12x^2 + 21x$
$7x + 7$

Grade 11 Polynomials and Rational Expressions CCSS: HSN-CN.C.9

Jeremy is working with the Fundamental Theorem of Algebra, and thinks he's found an exception. Looking at

$f(x) = 4(x-1)^2$ , this will result in only one root,

$x=1$ . Therefore, despite this being a second degree polynomial, there is only one root. Is this correct?

Yes, this is a known exception.
No, this is not a polynomial, it is a quadratic function.
No, if the quadratic formula is used, the other root is found.
No, this root has multiplicity of 2, which means it counts as two roots.

Grade 9 Polynomials and Rational Expressions

Classify the following polynomial.

$2x^3 + 8x^2 + 9x + 4$

Quartic trinomial
Cubic binomial
Quadratic trinomial
Cubic polynomial

Grade 7 Polynomials and Rational Expressions CCSS: 7.EE.A.1

Factor fully:

$6ab - 24bx =$ 6b(a - 4x)

Grade 11 Polynomials and Rational Expressions CCSS: HSN-CN.C.9

Looking at the graph of a quadratic polynomial, roots or zeros correspond to where the graph crosses the x-axis. When the graph just touches the x-axis, this corresponds to a double root. The Fundamental Theorem of Algebra states that a quadratic polynomial will always have 2 roots. How is this reconciled with a quadratic polynomial whose graph does not intersect the x-axis?

Quadratic polynomials always intersect the x-axis.
If a quadratic polynomial doesn't cross the x-axis it is no longer a polynomial, and the Fundamental Theorem of Algebra no longer applies.
When a quadratic polynomial doesn't cross the x-axis, this simply implies that its roots are complex with non-zero imaginary parts.
Simply translate the quadratic polynomial till it does cross the x-axis.

Grade 9 Polynomials and Rational Expressions

What is the degree of the following polynomial?

$3x^2y^4 - 2x^3y^5$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

Subtract the polynomials.

$(6x^3 - 4x^2y + 9xy^2 - 2y^3) - (xy^2 - 3x^2y + 4x^3 - 7y^3)$

$5x^3 - x^2y + 5xy^2 + 5y^3$
$2x^3 - x^2y + 8xy^2 +5y^3$
$2x^3 - 7x^2y + 8xy^2 - 9y^3$
$5x^3 -5x^2y + 12xy^2 - 9y^3$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-REI.A.1

Is the solution to the equation correct? If not, identify which step contains an error and why this step is incorrect.

$5x - 7 = 8$

Step 1:	$5x-7+7$	$= \ \ 8$
Step 2:	$5x$	$= \ \ 8$
Step 3:	$\(5x)/5$	$= \ \ 8/5$
Step 4:	$x$	$= \ \ 8/5$

It is correct.
Step 1 is incorrect because you should divide by 5 first.
Step 1 is incorrect because you have to add 7 to both sides.
Step 3 is incorrect because you only need to cancel the 5, not divide.

Grade 11 Polynomials and Rational Expressions

Simplify.

$(a - 1)/[(a)^2-1]$

$1/a, \ a!=0$
$1/(a - 1), \ a!= -1$
$1/(1- a), \ a!=1$
$1/(a + 1), \ a!=1$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-REI.A.1

A solution to the equation is given, except for one step. Choose the correct action for the missing step.

$2x + 3 = 4x$

Step 1:	$\ \ \$	$\ \ \$
Step 2:	$3$	$= \ \ 2x$
Step 3:	$3/2$	$= \ \ (2x)/2$
Step 4:	$3/2$	$= \ \ x$

Subtract 3 from both sides.
Add 3 to both sides.
Subtract 2x from both sides.
Divide by 2x on both sides.

Grade 10 Polynomials and Rational Expressions CCSS: HSA-SSE.A.2

Factor the expression.

$x^4 - y^4$

$(x - y)^4$
$(x^2 + y^2)(x + y)(x - y)$
$(x^2 + y^2)(x^2 - y^2)$
Can't be factored

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

$(2x+3)(5x-8)$

$10x^2+x-24$
$10x^2-31x-24$
$10x^2-x-24$
$7x^2+10x-5$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

$(3x^2y^3+xy^2)(x^4y-2x^2y^2)$

$3x^4y^2-6x^4y^6+x^5y^2-2x^3y^4$
$3x^6y^4-6x^4y^5+x^5y^3-2x^3y^4$
$3x^6y^4-6x^4y^5+x^4y^4-2x^2y^2$
$3x^4y^6-6x^5y^4+x^3y^5-2x^4y^3$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

$(10p^4+p^3-4p^2-8)-(-12p^4+5p^3-2p+3)$

$22p^4-4p^3-4p^2+2p-5$
$-2p^4-4p^3-4p^2+2p-5$
$-2p^4+6p^3-6p^2+2p-11$
$22p^4-4p^3-4p^2+2p-11$

Grade 9 Polynomials and Rational Expressions CCSS: HSA-APR.A.1

Add.

$(-4x^2 + 7x -11) + (6x^3 -7x +10)$

$2x^2 + x -1$
$6x^3 - 4x^2 -1$
$-2x^3 - x + 1$
$10x^2 + 14x + 21$

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