5 2factoring trinomial i

Factoring Trinomials I
Frank Ma © 2011

Trinomials (three-term) are polynomials of the form
ax2 + bx + c where a, b, and c are numbers.

ax2 + bx + c where a, b, and c are numbers. The product of two binomials is a trinomials:
(#x + #)(#x + #)  ax2 + bx + c

(#x + #)(#x + #)  ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a product of two binomials, if possible,

(#x + #)(#x + #)  ax2 + bx + c
Hence, to factor a trinomial, we write the trinomial as a product of two binomials, if possible, that is:
ax2 + bx + c  (#x + #)(#x + #)

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
We start with the case where a = 1, or trinomials
of the form x2 + bx + c.

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
To factorx2 + bx + c, we note hat
if (x + u)(x + v)

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv
= x2 + (u + v)x + uv

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c,

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c, and u + v = b.

(#x + #)(#x + #)  ax2 + bx + c
ax2 + bx + c  (#x + #)(#x + #)
if (x + u)(x + v)
= x2+ux + vx + uv
= x2 + (u + v)x + uv
= x2 + bx + c,
we need to have u and v where uv = c, and u + v = b. If this can’t be done, then the trinomial is prime (not factorable).

Example A.
a. Factor x2 + 5x + 6

Example A.
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v where uv = 6

Example A.
We want (x + u)(x + v) = x2+ 5x + 6, so we need u and v where uv = 6 and u + v = 5.

Example A.
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3)

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5,

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
b. Factor x2 – 5x + 6

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v where uv = 6

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
b. Factor x2– 5x + 6
We want (x + u)(x + v) = x2– 5x + 6, so we need u and v where uv = 6 and u + v = –5.

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).
c. Factor x2 + 5x – 6

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).
We want (x + u)(x + v) = x2+ 5x – 6, so we need uv = –6 and u + v = 5.

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,

Example A.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2 + 3 = 5, so x2 + 5x + 6 = (x + 2)(x + 3).
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5, so x2 – 5x + 6 = (x – 2)(x – 3).
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5, so x2 + 5x – 6 = (x – 1)(x + 6).

Observations About Signs

Given thatx2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.

1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.

In particular,
From the examples above
x2+ 5x + 6 = (x + 2)(x + 3)

In particular,
if b is negative, then both are negative.
x2+ 5x + 6 = (x + 2)(x + 3)

In particular,
x2+ 5x + 6 = (x + 2)(x + 3)
x2– 5x + 6 = (x – 2)(x – 3)
{

In particular,
x2+ 5x + 6 = (x + 2)(x + 3)
x2– 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs.
{

In particular,
x2+ 5x + 6 = (x + 2)(x + 3)
x2– 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
{

Given thatx2 + bx + c = (x + u)(x + v) so thatuv = c,
In particular,
x2+ 5x + 6 = (x + 2)(x + 3)
x2– 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
From the example above
x2– 5x – 6 = (x – 6)(x + 1)
{

Example B.
a. Factor x2 + 4x – 12

Example B.
We need u and v having opposite signs such that uv = –12,

Example B.
We need u and v having opposite signs such that uv = –12, u + v = +4.

Example B.
We need u and v having opposite signs such that uv = –12, u + v = +4. Since -12 = (-1)(12) = (-2)(6) = (-3)(4)…

Example B.
They must be –2 and 6

Example B.
They must be –2 and 6 hence x2 + 4x – 12 = (x – 2)(x + 6).

Example B.
b. Factor x2 – 8x – 12

Example B.
We need u and v such that uv = –12, u + v = –8 with
u and v having opposite signs.

Example B.
u and v having opposite signs. This is impossible.

Example B.
u and v having opposite signs. This is impossible.
Hence x2 – 8x – 12 is prime.

5 2factoring trinomial i

by math123a on Jun 22, 2011

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5 2factoring trinomial i — Presentation Transcript