Factoring Polynomials – Example

Example 1 Factor out the greatest common factor from each of the following polynomials.

(a) 8x⁴ − 4x³ +10x²

(b) x³ y² + 3x⁴ y + 5x⁵ y³

Solution

(a) 8x⁴− 4x³ +10x²

First we will notice that we can factor a 2 out of every term. Also note that we can factor an x²out of every term. Here then is the factoring for this problem.

8x⁴ − 4x³ +10x² = 2x² (4x² − 2x + 5)

Note that we can always check our factoring by multiplying the terms back out to make sure we get the original polynomial.

(b) x³ y² + 3x⁴ y + 5x⁵ y³

In this case we have both x’s and y’s in the terms but that doesn’t change how the process works. Each term contains and x3 and a y so we can factor both of those out. Doing this gives,

x³ y² + 3x⁴ y + 5x⁵ y³ = x³ y ( y + 3x + 5x² y²).

Example 2 Factor by grouping each of the following.

(a) 3x² − 2x + 12x −8

(b) x⁴ + x − 2x³ − 2

(c) x⁵ −3x³ − 2x² + 6

Solution

(a) 3x² − 2x +12x −8

In this case we group the first two terms and the final two terms as shown here,

(3x² − 2x)+ (12x −8)

Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping.

Doing this gives,

3x² − 2x +12x −8 = x (3x − 2) + 4(3x − 2)

We can now see that we can factor out a common factor of 3x − 2 so let’s do that to the final factored form.

3x² − 2x +12x − 8 = (3x − 2)(x + 4)

And we’re done. That’s all that there is to factoring by grouping. Note again that this will not always work and sometimes the only way to know if it will work or not is to try it and see what you get.

(b) x⁴ + x − 2x³ − 2

In this case we will do the same initial step, but this time notice that both of the final two terms are negative so we’ll factor out a “-” as well when we group them. Doing this gives,

(x⁴ + x)− (2x³ + 2)

Again, we can always distribute the “-” back through the parenthesis to make sure we get the original polynomial.

At this point we can see that we can factor an x out of the first term and a 2 out of the second term. This gives,

x⁴ + x − 2x³ − 2 = x (x³ +1)− 2(x³ +1)

We now have a common factor that we can factor out to complete the problem.

x⁴ + x − 2x³ − 2 = (x³ +1)(x − 2)

(c) x⁵ −3x³ − 2x² + 6

This one also has a “-” in front of the third term as we saw in the last part. However, this time the fourth term has a “+” in front of it unlike the last part. We will still factor a “-” out when we group however to make sure that we don’t lose track of it. When we factor the “-” out notice that we needed to change the “+” on the fourth term to a “-”. Again, you can always check that this was done correctly by multiplying the “-” back through the parenthsis.

(x⁵ −3x³) − (2x² − 6)

Now that we’ve done a couple of these we won’t put the remaining details in and we’ll go straight to the final factoring.

x⁵ − 3x³ − 2x² + 6 = x³ (x² −3)− 2(x² − 3) = (x² −3)(x³ − 2).

Example 3 Factor each of the following polynomials.

(a) x² + 2x −15

(b) x² −10x + 24

(c) x² + 6x + 9

Solution

(a) x² + 2x −15

Okay since the first term is x2 we know that the factoring must take the form.

x² + 2x −15 = (x + )(x + )

We know that it will take this form because when we multiply the two linear terms the first term must be x2 and the only way to get that to show up is to multiply x by x. Therefore, the first term in each factor must be an x. To finish this we just need to determine the two numbers that need to go in the blank spots.

We can narrow down the possibilities considerably. Upon multiplying the two factors out these two numbers will need to multiply out to get -15. In other words these two numbers must be factors of -15. Here are all the possible ways to factor -15 using only integers.

(−1)(15) (1)(−15) (−3)(5) (3)(−5)

Now, we can just plug these in one after another and multiply out until we get the correct pair.

However, there is another trick that we can use here to help us out. The correct pair of numbers must add to get the coefficient of the x term. So, in this case the third pair of factors will add to “+2” and so that is the pair we are after.

Here is the factored form of the polynomial.

x² + 2x −15 = ( x −3)(x + 5)

Again, we can always check that we got the correct answer my doing a quick multiplication.

Note that the method we used here will only work if the coefficient of the x2 term is one. If it is anything else this won’t work and we really will be back to trial and error to get the correct factoring form.

(b) x² −10x + 24

Let’s write down the initial form again,

x² − 10x + 24 = (x + __)(x + __)

Now, we need two numbers that multiply to get 24 and add to get -10. It looks like -6 and -4 will do the trick and so the factored form of this polynomial is,

x² − 10x + 24 = (x − 4)(x − 6)

(c) x² + 6x + 9

Again, let’s start with the initial form,

x² + 6x + 9 = (x + __ )(x + __)

This time we need two numbers that multiply to get 9 and add to get 6. In this case 3 and 3 will be the correct pair of numbers. Don’t forget that the two numbers can be the same number on occasion as they are here.

Here is the factored form for this polynomial.

x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)²

Note as well that we further simplified the factoring to acknowledge that it is a perfect square.

You should always do this when it happens.

Example 4 Factor each of the following.

(a) x² − 20x + 100

(b) 25x² − 9

(c) 8x³ + 1

Solution

(a) x² − 20x + 100

In this case we’ve got three terms and it’s a quadratic polynomial. Notice as well that the constant is a perfect square and its square root is 10. Notice as well that 2(10)=20 and this is the coefficient of the x term. So, it looks like we’ve got the second special form above. The correct factoring of this polynomial is,

x² − 20x +100 = ( x −10)²

To be honest, it might have been easier to just use the general process for factoring quadratic polynomials in this case rather than checking that it was one of the special forms, but we did need to see one of them worked.

(b) 25x² − 9

In this case all that we need to notice is that we’ve got a difference of perfect squares,

25x² − 9 = (5x)² − (3)²

So, this must be the third special form above. Here is the correct factoring for this polynomial.

25x² − 9 = (5x + 3)(5x − 3)

(c) 8x³ +1

This problem is the sum of two perfect cubes,

8x³ +1 = (2x)³ + (1)³

and so we know that it is the fourth special form from above. Here is the factoring for this polynomial.

8x³ +1 = (2x +1)(4x² − 2x +1)

Example 5 Factor each of the following.

(a) 3x⁴ − 3x³ − 36x²

(b) x⁴ − 25

(c) x⁴ + x² − 20

Solution

(a) 3x⁴ − 3x³ − 36x²

In this case let’s notice that we can factor out a common factor of 3x² from all the terms so let’s do that first.

3x⁴ − 3x³ −36x² = 3x² (x² − x −12)

What is left is a quadratic that we can use the techniques from above to factor. Doing this gives us,

3x⁴ −3x³ − 36x²= 3x² (x − 4)(x + 3)

Don’t forget that the FIRST step to factoring should always be to factor out the greatest common factor. This can only help the process.

(b) x⁴ − 25

There is no greatest common factor here. However, notice that this is the difference of two perfect squares.

x⁴ − 25 = (x²)² − 5²

So, we can use the third special form from above.

x⁴ − 25 = (x² + 5)(x² − 5)

Neither of these can be further factored and so we are done. Note however, that often we will need to do some further factoring at this stage.

(c) x⁴ + x² − 20

Let’s start this off by working a factoring a different polynomial.

u² + u − 20 = (u − 4)(u + 5)

We used a different variable here since we’d already used x’s for the original polynomial.

So, why did we work this? Well notice that if we let u = x² then u² = (x²)² = x4 . We can then rewrite the original polynomial in terms of u’s as follows,

x⁴ + x² − 20 = u² + u − 20

and we know how to factor this! So factor the polynomial in u’s then back substitute using the fact that we know u = x².

x⁴ + x² – 20 = u² + u – 20

= (u – 4) (u + 5)

= (x² – 4) (x + 5)

Finally, notice that the first term will also factor since it is the difference of two perfect squares.

The correct factoring of this polynomial is then,

x⁴ + x² − 20 = (x − 2)(x + 2)(x² + 5)

Note that this converting to u first can be useful on occasion, however once you get used to these this is usually done in our heads.

Sumber Alfi Blog Januari 25, 2020 Admin Bandung Indonesia

Factoring Polynomials – Example

Sabtu, 25 Januari 2020

Example 1 Factor out the greatest common factor from each of the following polynomials.

(a) 8x⁴ − 4x³ +10x²

(b) x³ y² + 3x⁴ y + 5x⁵ y³

Solution

(a) 8x⁴− 4x³ +10x²

First we will notice that we can factor a 2 out of every term. Also note that we can factor an x²out of every term. Here then is the factoring for this problem.

8x⁴ − 4x³ +10x² = 2x² (4x² − 2x + 5)

Note that we can always check our factoring by multiplying the terms back out to make sure we get the original polynomial.

(b) x³ y² + 3x⁴ y + 5x⁵ y³

In this case we have both x’s and y’s in the terms but that doesn’t change how the process works. Each term contains and x3 and a y so we can factor both of those out. Doing this gives,

x³ y² + 3x⁴ y + 5x⁵ y³ = x³ y ( y + 3x + 5x² y²).

Example 2 Factor by grouping each of the following.

(a) 3x² − 2x + 12x −8

(b) x⁴ + x − 2x³ − 2

(c) x⁵ −3x³ − 2x² + 6

Solution

(a) 3x² − 2x +12x −8

In this case we group the first two terms and the final two terms as shown here,

(3x² − 2x)+ (12x −8)

Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping.

Doing this gives,

3x² − 2x +12x −8 = x (3x − 2) + 4(3x − 2)

We can now see that we can factor out a common factor of 3x − 2 so let’s do that to the final factored form.

3x² − 2x +12x − 8 = (3x − 2)(x + 4)

(b) x⁴ + x − 2x³ − 2

In this case we will do the same initial step, but this time notice that both of the final two terms are negative so we’ll factor out a “-” as well when we group them. Doing this gives,

(x⁴ + x)− (2x³ + 2)

Again, we can always distribute the “-” back through the parenthesis to make sure we get the original polynomial.

At this point we can see that we can factor an x out of the first term and a 2 out of the second term. This gives,

x⁴ + x − 2x³ − 2 = x (x³ +1)− 2(x³ +1)

We now have a common factor that we can factor out to complete the problem.

x⁴ + x − 2x³ − 2 = (x³ +1)(x − 2)

(c) x⁵ −3x³ − 2x² + 6

(x⁵ −3x³) − (2x² − 6)

Now that we’ve done a couple of these we won’t put the remaining details in and we’ll go straight to the final factoring.

x⁵ − 3x³ − 2x² + 6 = x³ (x² −3)− 2(x² − 3) = (x² −3)(x³ − 2).

Example 3 Factor each of the following polynomials.

(a) x² + 2x −15

(b) x² −10x + 24

(c) x² + 6x + 9

Solution

(a) x² + 2x −15

Okay since the first term is x2 we know that the factoring must take the form.

x² + 2x −15 = (x + )(x + )

(−1)(15) (1)(−15) (−3)(5) (3)(−5)

Now, we can just plug these in one after another and multiply out until we get the correct pair.

Here is the factored form of the polynomial.

x² + 2x −15 = ( x −3)(x + 5)

Again, we can always check that we got the correct answer my doing a quick multiplication.

(b) x² −10x + 24

Let’s write down the initial form again,

x² − 10x + 24 = (x + __)(x + __)

Now, we need two numbers that multiply to get 24 and add to get -10. It looks like -6 and -4 will do the trick and so the factored form of this polynomial is,

x² − 10x + 24 = (x − 4)(x − 6)

(c) x² + 6x + 9

Again, let’s start with the initial form,

x² + 6x + 9 = (x + __ )(x + __)

Here is the factored form for this polynomial.

x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)²

Note as well that we further simplified the factoring to acknowledge that it is a perfect square.

You should always do this when it happens.

Example 4 Factor each of the following.

(a) x² − 20x + 100

(b) 25x² − 9

(c) 8x³ + 1

Solution

(a) x² − 20x + 100

x² − 20x +100 = ( x −10)²

(b) 25x² − 9

In this case all that we need to notice is that we’ve got a difference of perfect squares,

25x² − 9 = (5x)² − (3)²

So, this must be the third special form above. Here is the correct factoring for this polynomial.

25x² − 9 = (5x + 3)(5x − 3)

(c) 8x³ +1

This problem is the sum of two perfect cubes,

8x³ +1 = (2x)³ + (1)³

and so we know that it is the fourth special form from above. Here is the factoring for this polynomial.

8x³ +1 = (2x +1)(4x² − 2x +1)

Example 5 Factor each of the following.

(a) 3x⁴ − 3x³ − 36x²

(b) x⁴ − 25

(c) x⁴ + x² − 20

Solution

(a) 3x⁴ − 3x³ − 36x²

In this case let’s notice that we can factor out a common factor of 3x² from all the terms so let’s do that first.

3x⁴ − 3x³ −36x² = 3x² (x² − x −12)

What is left is a quadratic that we can use the techniques from above to factor. Doing this gives us,

3x⁴ −3x³ − 36x²= 3x² (x − 4)(x + 3)

Don’t forget that the FIRST step to factoring should always be to factor out the greatest common factor. This can only help the process.

(b) x⁴ − 25

There is no greatest common factor here. However, notice that this is the difference of two perfect squares.

x⁴ − 25 = (x²)² − 5²

So, we can use the third special form from above.

x⁴ − 25 = (x² + 5)(x² − 5)

Neither of these can be further factored and so we are done. Note however, that often we will need to do some further factoring at this stage.

(c) x⁴ + x² − 20

Let’s start this off by working a factoring a different polynomial.

u² + u − 20 = (u − 4)(u + 5)

We used a different variable here since we’d already used x’s for the original polynomial.

So, why did we work this? Well notice that if we let u = x² then u² = (x²)² = x4 . We can then rewrite the original polynomial in terms of u’s as follows,

x⁴ + x² − 20 = u² + u − 20

and we know how to factor this! So factor the polynomial in u’s then back substitute using the fact that we know u = x².

x⁴ + x² – 20 = u² + u – 20

= (u – 4) (u + 5)

= (x² – 4) (x + 5)

Finally, notice that the first term will also factor since it is the difference of two perfect squares.

The correct factoring of this polynomial is then,

x⁴ + x² − 20 = (x − 2)(x + 2)(x² + 5)

Note that this converting to u first can be useful on occasion, however once you get used to these this is usually done in our heads.

Sumber

Labels: Mathematician

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Factoring Polynomials – Example

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