Hints for solving the problems

Hints for solving the problems(7.2)

Factoring Polynomials II

Trinomials I

These algebraic expressions will have three terms. Following is a brief definition for factoring trinomials with the numerical coefficient of the first term (a) 1.

Remember a trinomial product has a general term like this:

ax² + bx + c

a is the numerical coefficient of the first term (quadratic term).

b is the numerical coefficient of the second term (linear term).

c is the third term or constant.

The following are the four types of trinomial product

Notice the signs of the second and third terms.

x² + 5x + 6 x² - 5x + 6 x² + x - 6 x² - x - 6

1·6 1·6 1·6 1·6

2·3 2·3 2·3 2·3

Write down all the factors of the third term (6). When the third term is positive the factors are added to get the numerical coefficient (b) of the middle term. When the third term is negative the factors are subtracted to get the numerical coefficient (b) of the middle term.

First trinomial 1 + 6 = 7, 2 + 3 = 5

So we use 2 & 3. Set up two binomials. When the third term is positive, both binomials will have the same sign as the middle term. When the third term is negative, the binomials will have different signs. (+ or -)

First trinomial: (x + 2)(x + 3) Check it and it multiplies to get the trinomial.

Second trinomial 1 + 6 = 7, 2 + 3 = 5

We use -2 and -3 because of the -5 in the middle.

Second trinomial (x - 2)(x - 3) It doesn’t make any difference whether the (x - 2) or the (x - 3) comes first.

Third trinomial 1 - 6 = -5, -1 + 6 = 5, 2 - 3 = -1, -2 + 3 = 1 Notice all combinations were subtracted and there is an x, which means 1x, in the middle term. When the third term is negative you subtract the factors of the third term to get the numerical coefficient (b) of the middle term.

The two factors to use would be -2 & 3. Why? -2 & 3 multiplied together gets -6 and -2 & 3 added together get 1

Answer is: (x + 3)(x - 2) Usually the largest factor takes the sign of the middle term.

Fourth trinomial 1 - 6 = -5, -1 + 6 = 5, 2 - 3 = -1, -2 + 3 = 1 The middle term has –x, which means -1x, so you use 2 & (-3).

Answer: (x + 2)(x - 3) Notice the largest factor (d) took the negative from the middle term.

Problem No. 7 - Factor: x² + 7x + 12 Remember from above, think of or write down all the factors of 12. 1, 12 & 2, 6 & 3,4

Which two factors added together (because the third term is positive) equals 7? It is the 3 & 4. Why? Because, 3 & 4 multiply together to get 12 and add together to get 7.

Answer: (x + 3)(x + 4) It doesn’t make any difference whether the (x + 4) or the (x + 3) is first.

Problem No. 16 - Factor: x² - 5x - 14 The third term is negative, so you have to subtract the factors of 14 to get -5 from the middle.

1 - 14 = -13, -1 + 14 = 13, 2 - 7 = -5, -2 + 7 = 5

Notice when subtracting, the factors change signs, so there is a positive and a negative answer.

From the problem you need a -5, so you will use the 2 & -7. Why? Because, 2 & -7 multiplied together will get -14, and added together will get -5.

Answer: (x + 2)(x - 7) It makes no difference which binomial comes first. I like to put the positive first and the negative second.

Problem No. 27 - Factor: x² - 7x - 60 Think of the factors of 60 that subtract to get 7.

1, 60, 2, 30, 3, 20, 4, 15, 5, 12, 6, 10

It is 5, 12. Think how to subtract the two factors, so the difference will be -7. 5 + (-12) = -7

Answer is: (x + 5)(x - 12)

Trinomials II

The problems become a little harder as the numerical coefficient of the quadratic term (a) will be greater than 1.

What you have to do here is take the factors of the first term (a) and the factors of the third term (c), multiply them together in such a way to be able to add or subtract the products to get the numerical coefficient of the middle term (b).

Problem No. 29 - Factor: 2x² + 7x + 5

1 2 1 5

You must add the cross-products to get 7, because of the + sign in front of the third term

So: (1 times 5) + (2 times 1) = 7

(1 times 1) + (2 times 5) = 11

(1 times 5) + (2 times 1) = 7 is the combination we use.

Now comes the problem of placing them in the correct spot in the binomials.

Remember the foil: inner two and outer two.

( x + )( x + ) Place the factors of the third term first in either place at the end of the binomials.

( x + 1)( x + 5) Now place the 2 in front of an x, but it must be in front of an x that will not be multiplied by the 5. Remember the outer and inner products were 5 & 2.

(x + 1)(2x + 5) Answer: Place the 2 in front of the x in the same binomial as the 5. This way because of the outer and inner products, they will not be multiplied together.

Problem No. 43 - Factor: 6x² - 11x - 10

Lets put down the factors: 1 6 1 10

2 3 2 5

Can you think of the factors of 6 multiplied by the factors of 10 and subtract (third term has a negative sign in front of it) to get 11.

1 6 & 1 10 (1 times 1) - (6 times 10) = -59

(1 times 10) - (6 times 1) = 4

1 6 & 2 5 (1 times 5) - (6 times 2) = -7

(1 times 2) - (6 times 5) = -28

I’m looking for -11

2 3 & 1 10 (2 times 1) - (3 times 10) = -28

(2 times 10) - (3 times 1) = 17

2 3 & 2 5 (2 times 5) - (3 times 2) = 4

(2 times 2) - (3 times 5) = -11

There it is, you use 2 3 & 2 5 Did you notice 4 and -15 are the cross products.

Now place the factors of the third term in the binomials first.

( x + 2)( x - 5) The 2 or the 5 take the negative and since -15 and 4 added together give you -11 the 5 takes the negative sign as it is a factor of -15.

When you place the factors of 6 make sure the 3 is multiplied by the -5 to get -15. According to the foil method the 3 must be in the first binomial with an x and the 2 will be in the second binomial with an x. Such as: (3x + 2)(2x - 5) the ans.

Make sure you use the foil method to check you answer. You must get the original trinomial or your answer is wrong.

Problem No. 55 - Factor: 9x² - 3xy - 2y²

Lets put down the factors: 1 9 1 2

3 3

Can you see the factors of 9 and the factors of 2, multiplied together, with the cross products subtracted to get -3. It is a good idea to start with the bottom two on each group.

I see it: 3 3 & 1 2 (3 times 2) - (3 times 1) = -3

Set up your binomials. ( x + 2)( x - 1) (3x + 2)(3x - 1)

This is wrong as when you multiply the two binomials together, the wrong sign comes out in the middle term, so change the signs of the binomials. (3x - 2)(3x + 1) the ans.

If you read the book you will notice some problem that say solve. To solve a quadratic equation, set the equation equal to zero. Move all the terms to the left or right (try to get the quadratic term positive) and then factor. This is putting the equation in standard form. Following is an example.

x² - 5x + 6 = 0 (x - 3)(x - 2) = 0 The Algebra behind this is, when factors are set to zero, each factor can be set to zero and solved as a linear equation. (a)(b) = 0, then a = 0 and b = 0.

(x - 3)(x - 2) = 0 x - 3 = 0 and x - 2 = 0. Solving for x you get x = 3 and x = 2. So the answer is: {2, 3}

Another example: Solve: x² - 4x = 0

Factor: x(x - 4) = 0 Set each factor equal to zero.

x = 0 and x - 4 = 0

Solving x = 0 and x = 4 The answer is: {0, 4}

Another example: Solve: 6x² = 63 - 13x

Put the equation in standard form, (move all terms to the left side of the equation and don’t forget to change the signs).

6x² + 13x - 63 = 0 Factor: (3x - 7)(2x + 9) = 0

Setting the factors = to 0 3x - 7 = 0 and 2x + 9 = 0

The answers are: {7/3, -9/2}