Exercise 137 Page 314

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.

4k^2+14k+12

SplitIntoFactors

Split into factors

2(2k^2)+ 2(7x)+ 2(6)

FactorOut

Factor out 5

2(2x^2+7x+6)

Factor the Quadratic Trinomial

Here we have a quadratic trinomial of the form ak^2+bk+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( 2k^2+7k+6 ) ⇕ 2( 2k^2+7k+6 ) We have that a= 2, b=7, and c=6. There are now three steps we need to follow in order to rewrite the above expression.

Find a c. Since we have that a= 2 and c=6, the value of a c is 2* (6)=12.
Find factors of a c. Since ac=12, which is positive, we need factors of a c to have the same signs in order for the product to be positive. Since b=7, which is also positive, the absolute value of the negative factor will need to be greater than the absolute value of the positive factor.

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

Rewrite bk as two terms. Now that we know which factors are the ones to be used, we can rewrite bk as two terms. 2(2k^2+7k+6 ) ⇕ 2( 2k^2+ 3x + 4k+6 )

Finally, we will factor the last expression obtained.

2(2k^2+3k+4k+6)

FactorOut

Factor out k

2(k(2k+3)+4k+6)

FactorOut

Factor out 2

2(k(2k+3)+2(2k+3))

FactorOut

Factor out (2k+3)

2(2k+3)(k+2)

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

Factor the Quadratic Binomial

Look closely at the expression x^2-25. It can be expressed as the difference of two perfect squares.

x^2-25

WritePow

Write as a power

x^2-5^2

Recall the formula to factor a difference of squares. a^2- b^2 ⇔ ( a+ b)( a- b) We can apply this formula to our expression. 3( x^2- 5^2) ⇔ 3( x+ 5)( x- 5)

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

2x^3y+8x^2y+6xy^2

SplitIntoFactors

Split into factors

2xy(x^2)+ 2xy(4x)+ 2xy(3y)

FactorOut

Factor out 2xy

2xy(x^2+4x+3y)

Factor the Quadratic Trinomial

Let's look closely at the trinomial between the parentheses. x^2+4x+3y Notice that we have one x^2-term, one x-term, and one y-term. If this expression could be factored further, we would have two factors, each with x within them. However, that way we would also have one xy-term which is not present. (ax+b)(cx+dy) ⇕ acx^2 + ad xy + bcx + bd y In our expression the terms containing x^2 and y are both non-zero, so neither a nor d can be equal to 0. Therefore, their product must also be non-zero. As such, since there is no xy-term in our trinomial, the expression cannot be factored any further. 2x^3y+8x^2y+6xy^2 ⇓ 2xy(x^2+4x+3y)