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Algebra II : Polynomials

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Example Questions

Example Question #5 : Polynomials

Factor completely: $x^{2} - 9x - 20$

Possible Answers:

(x+10)(x-2)

(x-10)(x+2)

(x+5)(x-4)

The polynomial cannot be factored further.

(x-5)(x+4)

Correct answer:

The polynomial cannot be factored further.

Explanation:

We are looking to factor this quadratic trinomial into two factors (x+ ? )(x+?) , where the question marks are to be replaced by two integers whose product is and whose sum is .

We need to look at the factor pairs of in which the negative number has the greater absolute value, and see which one has sum :

$\begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-20& -19\\ 2,-10&-8 \\ 4,-5& -1 \end{matrix}$

None of these pairs have the desired sum, so the polynomial is prime.

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Example Question #11 : Polynomials

Factor completely: $3y^{2} + y - 10$

Possible Answers:

$\left (y+ 2 \right ) \left (3y -5 \right )$

$\left (y- 2 \right ) \left (3y +5 \right )$

The polynomial cannot be factored further.

$\left (y+ 10 \right ) \left (3y -1\right )$

$\left (y- 10 \right ) \left (3y +1\right )$

Correct answer:

$\left (y+ 2 \right ) \left (3y -5 \right )$

Explanation:

Rewrite this as $3y^{2} + 1 y - 10$

Use the -method by splitting the middle term into two terms, finding two integers whose sum is 1 and whose product is 3 (-10) = -30 ; these integers are -5,6 , so rewrite this trinomial as follows:

$3y^{2} + 1 y - 10 = 3y^{2} -5y + 6 y - 10$

Now, use grouping to factor this:

$\left (3y^{2} -5y \right )+ \left (6 y - 10 \right )$

$= y \left (3y -5 \right )+ 2 \left (3y -5 \right )$

$= \left (y+ 2 \right ) \left (3y -5 \right )$

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Example Question #34 : Factoring Polynomials

Factor the expression:

$\small 49r^2 - 36$

Possible Answers:

$\small \small (7r+6)(7r-6)$

$\small (7r^2+6)(7r^2-6)$

$\small prime$

$\small (7r-6)^2$

$\small (7r^2-6)$

Correct answer:

$\small \small (7r+6)(7r-6)$

Explanation:

The given expression is a special binomial, known as the "difference of squares". A difference of squares binomial has the given factorization: $\small a^2-b^2=(a+b)(a-b)$ . Thus, we can rewrite $\small 49r^2 - 36$ as $\small (7r)^2 - (6)^2$ and it follows that $\small (7r)^2-(6)^2=(7r+6)(7r-6).$

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Example Question #2 : Equations / Solution Sets

Factor the equation:

$x^{2}-6x+8$

Possible Answers:

(x-4)(x-2)

(x+4)(x+2)

(x-8)(x+2)

(x-8)(x-2)

(x+4)(x-2)

Correct answer:

(x-4)(x-2)

Explanation:

The product of (x+a)(x+b) is $x^{2}+ax+bx+ab$ .

For the equation $x^{2}-6x+8$ ,

a+b must equal and must equal .

Thus and must be and , making the answer (x-4)(x-2) .

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Example Question #81 : Polynomials

Find solutions to $x^{2}+x-6 = 0$ .

Possible Answers:

x=-2,4

x=-3,2

x=-1,6

x=2,3

x=-6,1

Correct answer:

x=-3,2

Explanation:

The quadratic can be solved as (x-2)(x+3)=0 . Setting each factor to zero yields the answers.

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Example Question #42 : Factoring Polynomials

Factor:

x^2-25

Possible Answers:

x(x-25)

(x-5)^2

(x-5)(x+5)

The expression cannot be factored.

(x+5)^2

Correct answer:

(x-5)(x+5)

Explanation:

Because both terms are perfect squares, this is a difference of squares:

$\small x^2-25=x^2-5^2$

The difference of squares formula is a^2-b^2 = (a+b)(a-b) .

Here, a = x and b = 5. Therefore the answer is $\small (x-5)(x+5)$ .

You can double check the answer using the FOIL method:

$\small x^2-5x+5x-25= x^2-25$

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Example Question #1221 : Algebra Ii

Factor:

$\small x^2-5x+4$

Possible Answers:

$\small (x-1)(x+4)$

$\small (x-1)(x-4)$

$\small (x+1)(x+4)$

$\small (x+1)(x-4)$

Correct answer:

$\small (x-1)(x-4)$

Explanation:

The solutions indicate that the answer is:

$\small (x\ 1)(x\ 4)$ and we need to insert the correct addition or subtraction signs. Because the last term in the problem is positive (+4), both signs have to be plus signs or both signs have to be minus signs. Because the second term (-5x) is negative, we can conclude that both have to be minus signs leaving us with:

$\small (x-1)(x-4)$

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Example Question #41 : Factoring Polynomials

Factor the following polynomial: x^2-2x-35 .

Possible Answers:

(x+5)(x-7)

(x+5)(x+7)

(x-1)(x+35)

(x-5)(x-7)

(x-35)(x-1)

Correct answer:

(x+5)(x-7)

Explanation:

Because the x^2 term doesn’t have a coefficient, you want to begin by looking at the term ( ax^2+bx+c ) of the polynomial: . Find the factors of that when added together equal the second coefficient (the term) of the polynomial.

There are only four factors of : 1, 5, 7, 35 , and only two of those factors, 5, 7 , can be manipulated to equal when added together and manipulated to equal when multiplied together: 5, -7 (i.e., 5-7=-2 ).

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Example Question #11 : How To Factor A Variable

Factor the following polynomial: x^2+13x+30 .

Possible Answers:

(x-10)(x-3)

(x-5)(x-6)

(x+5)(x+6)

(x-10)(x+3)

(x+10)(x+3)

Correct answer:

(x+10)(x+3)

Explanation:

Because the x^2 term doesn’t have a coefficient, you want to begin by looking at the term ( ax^2+bx+c ) of the polynomial: .

Find the factors of that when added together equal the second coefficient (the term) of the polynomial: .

There are seven factors of : 1, 2, 3, 5, 6, 10, 15 , and only two of those factors, 3, 10 , can be manipulated to equal when added together and manipulated to equal when multiplied together: 3, 10

$3+10=13, 3\times10=30$

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Example Question #1 : How To Find The Degree Of A Polynomial

Simplify:

$\frac{2x-5}{5-2x}$

Possible Answers:

None of the above

-1

Correct answer:

-1

Explanation:

The given expression can be re-written as:

$\frac{2x-5}{\left -(2x-5 \right )}$

Cancel (2x - 5):

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

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Algebra II : Polynomials

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #5 : Polynomials

Example Question #11 : Polynomials

Example Question #34 : Factoring Polynomials

Example Question #2 : Equations / Solution Sets

Example Question #81 : Polynomials

Example Question #42 : Factoring Polynomials

Example Question #1221 : Algebra Ii

Example Question #41 : Factoring Polynomials

Example Question #11 : How To Factor A Variable

Example Question #1 : How To Find The Degree Of A Polynomial

All Algebra II Resources

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