6.4 Factoring and Solving Polynomial

Expressions

p. 345

Types of Factoring:

• From Chapter 5 we did factoring of:– GCF : 6x2 + 15x = 3x (2x + 5)– PTS : x2 + 10x + 25 = (x + 5)2

– DOS : 4x2 – 9 = (2x + 3)(2x – 3)– Bustin’ da B = 2x2 – 5x – 12 =

» (2x2 - 8x) + (3x – 12) =» 2x(x – 4) + 3(x – 4)=» (x – 4)(2x + 3)

Now we will use Sum of Cubes:

• a3 + b3 = (a + b)(a2 – ab + b2)

• x3 + 8 =

• (x)3 + (2)3 =

• (x + 2)(x2 – 2x + 4)

Difference of Cubes

• a3 – b3 = (a – b)(a2 + ab + b2)

• 8x3 – 1 =

• (2x)3 – 13 =

• (2x – 1)((2x)2 + 2x*1 + 12)

• (2x – 1)(4x2 + 2x + 1)

When there are more than 3 terms – use GROUPING

• x3 – 2x2 – 9x + 18 =

• (x3 – 2x2) + (-9x + 18) = Group in two’s

• with a ‘+’ in the middle

• x2(x – 2) - 9(x – 2) = GCF each group

• (x – 2)(x2 – 9) =

• (x – 2)(x + 3)(x – 3) Factor all that can be

• factored

Factoring in Quad form:

• 81x4 – 16 =

• (9x2)2 – 42 =

• (9x2 + 4)(9x2 – 4)= Can anything be

• factored still???

• (9x2 + 4)(3x – 2)(3x +2)

• Keep factoring ‘till you can’t factor any more!!

You try this one!

• 4x6 – 20x4 + 24x2 =

• 4x2 (x4 - 5x2 +6) =

• 4x2 (x2 – 2)(x2 – 3)

In Chapter 5, we used the zero property. (when

multiplying 2 numbers together to get 0 – one must

be zero)The also works with higher

degree polynomials

Solve:• 2x5 + 24x = 14x3

• 2x5 - 14x3 + 24x = 0 Put in standard form• 2x (x4 – 7x2 +12) = 0 GCF• 2x (x2 – 3)(x2 – 4) = 0 Bustin’ da ‘b’• 2x (x2 – 3)(x + 2)(x – 2) = 0 Factor • everything• 2x=0 x2-3=0 x+2=0 x-2=0 set all • factors to 0• X=0 x=±√3 x=-2 x=2

Now, you try one!

• 2y5 – 18y = 0

• Y=0 y=±√3 y=±i√3

Assignment