Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
A zero (root) of a function is the x-intercept of the graph. Quadratic functions can have 0, 1, or 2 zeros. (In general, a function can have as many zeros as its highest exponent.)
The zeros of a quadratic function are always symmetric about the axis of symmetry.
Zeroes can be found by graphing or by factoring.No zeros: 1 zero: 2 zeros:
Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
Factoring by GCF
The GCF (Greatest Common Factor) is the greatest number and/or variable that evenly divides into each term.
Factor each expression by GCF:
1) 4xy2 – 3x 2) 10x2y3 – 20xy2 – 5xy
3) 3n4 + 6m2n3 – 12nm 4) 5x2 + 7
x(4y – 3x) 5xy(2xy2 – 4y – 1)
3n(n3 + 2m2n2 – 6m) Prime
Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
Determine the zeros of each function:
1)f(x) = 5x2 + 10x
2)g(x) = ½x2 – 2x
3)h(x) = 9x2 + 3x
5x(x + 2)5x = 0 and x + 2 = 0x = 0 and x = -2The zeros are x = 0 and x = -2
½x(x – 4)½x = 0 and x – 4 = 0x = 0 and x = 4The zeros are x = 0 and x = 4
3x(3x + 1)3x = 0 and 3x + 1 = 0x = 0 and 3x = -1 x = -1/3The zeros are x = 0 and x = -1/3
Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
A binomial (quadratic expression with two terms) consisting of two perfect squares can be factored using a method called “difference of squares.”
Difference of squares:a2 – b2 = (a + b)(a – b)
Ex) Factor each expression:1) x2 – 9 2) 16x2 – 49
(x + 3)(x – 3) (4x + 7)(4x – 7)
Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
Find the roots of the equation by factoring.
Example 4A: Find Roots by Using Special Factors
12x2 - 27
3(4x2 – 9)
3(2x – 3)(2x + 3) = 0
2x – 3 = 0 2x + 3 = 0
x = 3/2 x = -3/2
The zeros are x = 3/2 and x = -3/2
Holt Algebra 2
5-3 Solving Quadratic Equations by Graphing and Factoring
Factor by grouping when you have four terms with no common factor.
Ex) Factor each expression:1)xy – 5y – 2x + 10
2) x2 + 4x – x – 4
y(x – 5) – 2(x – 5)
(y – 2)(x – 5)
x(x + 4) – 1(x + 4)(x – 1)(x + 4)x – 1 = 0 x + 4 = 0x = 1 and x = -4The zeros are x = 1 and x = -4