Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 7.1
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Exercise 18 Page 379

Practice makes perfect
a We want to completely factor the given expression. To do so, we will first identify and factor out the greatest common factor.

Factor Out the GCF

The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 3.
15x^2+39x-18
3(5x^2)+ 3(13x)- 3(6)
3(5x^2+13x-6)

Factor the Quadratic Trinomial

Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 3( 5x^2+13x-6 ) ⇕ 3( 5x^2+13x+(- 6) ) We have that a= 5, b=13, and c=- 6. There are now three steps we need to follow in order to rewrite the above expression.

  1. Find a c. Since we have that a= 5 and c=- 6, the value of a c is 5* (- 6)=- 30.
  2. Find factors of a c. Since ac=- 30, which is negative, we need factors of a c to have opposite signs — one positive and one negative. Since b=13, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor, so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &30 &- 1 + 30 &29 - 2 & 15 & - 2 + 15 &13 - 3 &10 &- 3 + 10 &7 - 5 &6 &- 5 + 6 &1

  1. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3(5x^2+13x-6 ) ⇕ 3 ( 5x^2 - 2x + 15x-6 )
Finally, we will factor the last expression obtained.
3( 5x^2-2x+15x-6 )
3( x(5x-2)+15x-6 )
3( x(5x-2)+3(5x-2) )
3 ( (x+3)(5x-2) )
3(x+3)(5x-2)
b We want to completely factor the given expression. To do so, we will first identify and factor out the greatest common factor.

Factor Out the GCF

The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 2.
6t^2-26t+8
2(3t^2)- 2(13t)+ 2(4)
2(3t^2-13t+4)

Factor the Quadratic Trinomial

Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. 2( 3t^2-13t+4 ) ⇕ 2( 3t^2+(- 13)t+4 ) We have that a= 3, b=- 13, and c=4. There are now three steps we need to follow in order to rewrite the above expression.

  1. Find a c. Since we have that a= 3 and c=4, the value of a c is 3* (4)=12.
  2. Find factors of a c. Since a c=12, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=- 13, which is negative, those factors will need to be negative so that their sum is negative.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 & - 12 & - 1 + ( - 12) &- 13 - 2 &- 6 &- 2 + (- 6) &- 8 - 3 &- 4 &- 3 + (- 4) &- 7

  1. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 2(3t^2- 13t+4 ) ⇕ 2 ( 3t^2 - 1t - 12t+4 )
Finally, we will factor the last expression obtained.
2( 3t^2-1t-12t+4 )
2( t(3t-1)-12t+4 )
2( t(3t-1)-4(3t-1) )
2 ( (t-4)(3t-1) )
2(t-4)(3t-1)
c We want to factor the given equation. Let's start factoring by first identifying the greatest common factor (GCF).

Factor Out the GCF

The GCF of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. In this case, the GCF is 6.
6x^2-24
6* x^2- 6* 4
6(x^2-4)

The result of factoring out a GCF from the given expression is a binomial with a leading coefficient of 1. 6( x^2-4) Let's temporarily only focus on this binomial, and we will bring back the GCF after factoring.

Factor the Expression

Do you notice that the binomial is a difference of two perfect squares? This can be factored using the difference of squares method. a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.

Expression x^2-4
Rewrite as Perfect Squares x^2 - 2^2
Apply the Formula (x+2)(x-2)

Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 6(x+2)(x-2)