Algebra 1 Glencoe McGraw-Hill JoAnn Evans

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Presentation transcript:

Algebra 1 Glencoe McGraw-Hill JoAnn Evans Math 8H 8-4 (day 2) More Factoring Practice Algebra 1 Glencoe McGraw-Hill JoAnn Evans

3y(2x - 1)(4x + 3) is the factored form. 24x2y + 6xy – 9y Start every factoring problem by checking for a greatest common monomial factor. 3y(8x2 + 2x – 3) -24 Check your solution! ● 6 -4 + 2 3y(2x - 1)(4x + 3) is the factored form.

(2y + 5)(5y + 3) is the factored form. 150 ● Check your solution! 6 25 + 31 (2y + 5)(5y + 3) is the factored form.

(x + 6)(4x - 1) is the factored form. Put the polynomial in standard form before you begin to factor. 4x2 + 23x – 6 -24 ● 24 -1 + 23 (x + 6)(4x - 1) is the factored form.

Notice the power of the first term. Practice Problems 4 - 25n + 25n2 x4 - 10x2 + 25 25n2 - 25n + 4 Notice the power of the first term. 100 25 ● -5 ● -20 -5 -5 + + -25 -10 (5n - 1)(5n - 4) (x2 - 5)(x2 - 5)

Solve Equations by Factoring (find the roots) x2 = -8x Set the equation equal to 0. x2 + 8x = 0 Factor the polynomial completely. x(x + 8)= 0 Set each of the factors equal to 0. x = 0 or x + 8 = 0 x =

Solve Equations by Factoring (find the roots) 9y2 = 12y - 4 36 Set the equation equal to 0. Make sure it’s in standard form. ● -6 -6 + -12 9y2 – 12y + 4 = 0 Factor the polynomial completely. (3y – 2)(3y - 2)= 0 Set each of the factors equal to 0. 3y - 2 = 0

1st integer  2nd integer = 80 Find two consecutive positive even integers whose product is 80. Let x = the 1st integer Let x + 2 = the 2nd integer 1st integer  2nd integer = 80 x(x + 2) = 80 x2 + 2x = 80 -80 x2 + 2x – 80 = 0 ● 10 -8 + (x + 10) (x – 8) = 0 2 x + 10 = 0 or x – 8 = 0 x = -10 or 8 Remember the problem asked for positive consecutive even integers. Only one of the roots will fit the context of this problem. The 1st integer is 8 and the 2nd integer is 10.

The width of a rectangular poster is 7 in. less than its length The width of a rectangular poster is 7 in. less than its length. If the poster’s area is 78 in2, find the dimensions. Let x = length Let x - 7 = width x - 7 length  width = area x x(x - 7) = 78 -78 x2 - 7x = 78 ● 6 -13 + x2 - 7x – 78 = 0 -7 (x + 6) (x – 13) = 0 Only the positive root makes sense in this context. Length can’t be negative. Discard the negative root. x = -6 or 13 Length is 13 and width is 6.

square of 1st + square of 2nd = 145 The sum of the squares of two negative consecutive integers is 145. Find the integers. Let x = 1st integer Let x + 1 = 2nd integer square of 1st + square of 2nd = 145 x2 + (x + 1)2 = 145 -72 x2 + (x + 1)(x + 1) = 145 ● 9 -8 + x2 + x2 + x + x + 1 = 145 1 2x2 + 2x + 1 = 145 x = -9 or 8 2x2 + 2x – 144 = 0 Which answer doesn’t make sense in the context of this problem? 2(x2 + x – 72) = 0 2(x + 9)(x – 8) = 0 x + 9 = 0 or x – 8 = 0 The integers are -9 & -8.