9.5 Factoring trinomials x² + bx + c
Name: ______________________
(x + 3)(x + 4) =
We will reverse this process to factor trinomial of the form ax² + bx + c
Steps: 1. Look at the leading coefficient. Is it a 1?
2. List the multiples of the constant term (can do this in calculator).
3. Pick and circle the pair that adds to give the middle term.
o If constant term was positive (c > 0), you will have the same signs (two + OR two –);
o If constant term was negative (c < 0), you will have numbers with different signs.
4. Insert the pair into parentheses: (x + ___ )(x + ___ )
5. Check by multiplying the binomials and see if you get the original problem.
1. x2 + 8x + 7
2. b2 – 7b + 10
3. w2 – 12w – 13
4. –p2 – 10p – 25
5. –m2 + 10m – 24
6. y2 – 5y – 24
7. a2 + 13a + 36
8. –n2 – 2n + 48
9. –z2 + 14z – 40
Solve the equation.
Factor first, then set each factor equal to Zero and solve using Zero Product Property
10. y2 + 17y + 72 = 0
11. a2 – 9a – 36 = 0
12. –w2 + 13w – 42 = 0
13. m2 – 5m – 14 = 0
14. x2 + 11x + 24 = 0
15. –n2 +12n – 27 =0
Find the zeros of the polynomial function and sketch the parabolas
16. f(x) = x2 – 5x – 36
17. g(x) = –x2 – 8x + 20
18. h(x) = x2 – 11x + 24
19. f(x) = –x2 – 11x – 28
20. g(x) = x2 + 6x –9
21. h(x) = x2 + 5x + 6
Quadratic formula
x=
3x2 + 5x
a=
gives the solutions of any quadratic equation in standard form.
8=0
Write in standard form.
,b=
,c=
D=
.
Recognize the coefficients
=
Evaluate the discriminant. Stop if D < 0, the trinomial cannot be
factored
x=
The solutions are
x1 =
=
and
Factor: ax² + bx + c = a(x – x1) (x – x2)
3x2 + 5x 8 = 3(x –
)(x –
)=
x1 =
=
x2 =
=
x2 =
=