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College Algebra : Factoring Polynomials

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Example Question #11 : Factoring Polynomials

Factor the polynomial

$2X^{2}-9X-5=0$

Possible Answers:

(2x+1)(2x-5)=0

(2x+1)(x-5)=0

(2x+1)(x+5)=0

(2x-1)(x-5)=0

(x+1)(x-5)=0

Correct answer:

(2x+1)(x-5)=0

Explanation:

$2X^{2}-9X-5=0$ needs to be seen as $AX^{2}+BX+C=0$ . Then we need to ask what factors of "C" will equal "B" if one of them is multiplied by "A"

So first thing is to find factors of "C," luckily 5 only has 2:

1 and 5

We need to write an equation that uses either 5, 1 and a "*2" to equal -9.

(-5*2)+1=9

Now we can write out our factors knowing that we will be using -5,2,and 1.

(__X+___)(___X+____)

from the work above, we know we have to multiply the 2 and the -5, so they need to be in opposite factors

(2X+__)(X-5)

that only leaves one space for the 1

(2X+1)(X-5)

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Example Question #11 : Factoring Polynomials

Factor the polynomial

X^2+3X-18

Possible Answers:

(X-2)(X+9)

(X+3)(X+6)

(X-3)(X-6)

(X-3)(X+6)

(X+3)(X-6)

Correct answer:

(X-3)(X+6)

Explanation:

X^2+3X-18 needs to be seen as AX^2+BX+C=0 . Then you need to ask what factors of "C" will equal "B" when added together.

C=-18, factors of -18 are:

-1,18

-2,9

-3,6

1,-18

2,-9

3,-6

Of those factors, only -3,6 will give us "B", which in this case, is "3"

X^2+3X-18

becomes

(X-3)(X+6)

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Example Question #11 : Factoring Polynomials

Factor completely:

$x^{4}- 9x^{2}+ 8$

Possible Answers:

$(x-1) ^{2}(x^{2}+8)$

$(x+1)^{2}(x^{2}-8)$

$(x+1)(x-1) (x^{2}-8)$

$(x+1)(x-1) (x^{2}+8)$

$(x-1) ^{2}(x+2)(x+4)$

Correct answer:

$(x+1)(x-1) (x^{2}-8)$

Explanation:

Set $t = x^{2}$ , and, consequently, $t^{2} = (x^{2} )^{2}= x^{4}$ . Substitute to form a quadratic polynomial in :

$x^{4}- 9x^{2}+ 8$

$= t^{2}- 9t + 8$

Factor this trinomial by finding two numbers whose product is 8 and whose sum is . Through trial and error, these numbers can be found to be and , so

=(t-1)(t-8)

Substitute $x^{2}$ back for :

$=(x^{2}-1)(x^{2}-8)$

The first binomial is the difference of squares; the second is prime since 8 is not a perfect square. Thus, the final factorization is

$=(x+1)(x-1) (x^{2}-8)$

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Example Question #11 : Factoring Polynomials

Factor completely:

$x^{6}- 9x^{3}+ 8$

Possible Answers:

$(x -1)(x^{2}+x+1)(x -2)(x^{2}+2x+ 4)$

$(x +1)(x^{2}+x-1)(x +2)(x^{2}+2x-4)$

$(x -1)(x^{2} +1)(x -2)(x^{2} + 4)$

$(x -1)(x^{2}+x-1)(x -2)(x^{2}+x- 4)$

$(x -1)(x^{2}+x+1)(x -2)(x^{2}+x+ 4)$

Correct answer:

$(x -1)(x^{2}+x+1)(x -2)(x^{2}+2x+ 4)$

Explanation:

Set $t = x^{3}$ , and, consequently, $t^{2} = (x^{3} )^{2}= x^{6}$ . Substitute to form a quadratic polynomial in :

$x^{6}- 9x^{3}+ 8$

$= t^{2}- 9t + 8$

Factor this trinomial by finding two numbers whose product is 8 and whose sum is . Through trial and error, these numbers can be found to be and , so

=(t-1)(t-8)

Substitute $x^{3}$ back for :

$=(x^{3}-1)(x^{3}-8)$

Both factors are the difference of perfect cubes and can be factored further as such using the appropriate pattern:

$=(x -1)(x^{2}+x \cdot 1+1^{2})(x -2)(x^{2}+x \cdot 2+ 2^{2})$

$=(x -1)(x^{2}+x+1)(x -2)(x^{2}+2x+ 4)$

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Example Question #11 : Factoring Polynomials

Factor: x^2-8x-20

Possible Answers:

(x+10)(x-2)

-(x-10)(x+2)

(x-10)(x-2)

(x-10)(x+2)

Correct answer:

(x-10)(x+2)

Explanation:

Step 1: Break down into factors...

20: 1,2,4,5,10,20

Step 2: Find two numbers that add or subtract to .

We will choose and .

Step 3: Look at the equation and see which number needs to change sign..

According to the middle term, must be negative.

So, the factors are and .

Step 4: Factor by re-writing the solutions in the form:

(x-x_1)(x-x_2)

So... (x-10)(x+2)

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College Algebra : Factoring Polynomials

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #11 : Factoring Polynomials

Example Question #11 : Factoring Polynomials

Example Question #11 : Factoring Polynomials

Example Question #11 : Factoring Polynomials

Example Question #11 : Factoring Polynomials

All College Algebra Resources

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