Algebra Help Factoring Trinomials
Factoring trinomials (quadratic): clear explanation of the method with examples
Consider the product of two linear expressions (y + a) and (y + B).
(Y + a) (y + b) y = (y + b), + a + (b) = y ^ 2 + ab + ay + = y ^ 2 + y (a + b) ab +
We can write as
y ^ 2 + Y (a + b) ab + = (y + a) (y + b) …….( i)
Similarly, consider the product of two linear expressions (Ay + b) and (c + d).
(Ay + b) (c + d) y = (c + d) b + (c + d) = ^ 2 + acy BCY + ady + bd = acy ^ 2 + y (ad + bc) + bd
We can write as
acy ^ 2 + y (ad + bc) + Bd = (ay + b) (c + d) …….( ii)
Equation (i) is simple quadratic polynomial expressed as a product of two linear factors
and equation (Ii) is the general quadratic polynomial expressed as the product of two linear factors
Observation of the two formulations, leads to the method of factoring quadratic expressions.
In equation (i)
the product of the coefficient of ab y ^ 2 and the constant term =
and the coefficient of y = a + b = sum of factors of ab
Similarly, in equation (Ii)
the product of the coefficient of y ^ 2 and the constant term = (ac) (bd) = (ad) (bc)
and the coefficient of y = (ad + bc) = Sum of the factors acbd
So if we can solve the product of y ^ 2 and the constant term in the product of two factors so that their sum is equal the coefficient of y, then we can factor the quadratic expression.
We discuss the next steps in the method and apply it to solve a series problems.
Factoring trinomials method (quadratic):
Step 1:
Multiply the coefficient of y ^ 2 by the constant term.
Step 2:
Solve this product into two factors such that their sum is the coefficient y
Step 3:
Rewrite the term and the sum of two terms with these factors as coefficients.
Step 4:
Then take the common factor in the first two terms and conditions of the last two.
Step 5:
Then take the common factor of two terms thus formed.
What you get in step 5 is the product of two factors required.
The method is clear by the following worked examples.
The examples are chosen so that all models are covered.
Example 1:
Factoring 9y ^ 2 + 26y + 16
Solution:
Let P = 9y ^ 2 + 26y + 16
Now follow the five steps listed above.
Step 1:
(Coefficient of y ^ 2) x (constant term) = 9 x 16 = 144
Step 2:
We have to express 144 as two factors whose sum is the coefficient of x = 26;
144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18, (8 + 18 = 26)
Step 3:
P = 9y ^ 2 + 26y + 16 = 9y ^ 2 + 8y + 18Y + 16
Step 4:
P = y (9y + 
+ 2 (9y +
Step 5:
P = (9y + (y + 2)
Therefore, 9y ^ 2 + 26y + 16 = (9y + (y + 2) American National Standard.
Example 2:
Factor y ^ 2 + 7y – 78
Solution:
Let P = y ^ 2 + 7y – 78
Now, follow the five steps listed above.
Step 1:
(Coefficient of y ^ 2) x (constant term) = 1 x = -78 -78
Step 2:
We must express -78 As two factors whose sum is the coefficient of y = 7;
-78 = -2 X 39 x 3 = -2 x 13 x 13 = -6; (-6 + 13 = 7)
Step 3:
P = Y ^ 2 + 7y – 78 = y ^ 2 – 6y + 13 And – 78
Step 4:
P = y (y – 6) + 13 (y – 6)
Step 5:
P = (Y – 6) (y + 13)
Therefore, y ^ 2 + 7y – 78 = (y – 6) (y + 13) Response.
Example 3:
Factoring 4y ^ 2 – 5y + 1
Solution:
Let P = 4y ^ 2 – 5y + 1
Now, follow the five steps listed above.
Step 1:
(Coefficient of y ^ 2) x (constant term) = 4 x 1 = 4
Step 2:
We must express 4, two factors whose sum is the coefficient of y = -5;
4 = 4 x 1 x = -4 -1 [(-4) + (-1) = -5]
Step 3:
P = 4y ^ 2 – 5y + 1 = 4y ^ 2 – 4y – y + 1
Step 4:
P = 4y (y – 1) – 1 (y – 1)
Step 5:
P = (y – 1) (4y – 1)
Therefore, 4y ^ 2 – 5y + 1 = (y – 1) (4y – 1) Response.
Example 4:
Factor 3y ^ 2 – 17 and – 20
Solution:
Let P = 3y ^ 2 – 17 and – 20
Now, follow the five steps listed above.
Step 1:
Coefficient and constant term x ^ 2 = 3 x = -20 -60
Step 2:
We must express -60 as two factors whose sum = coefficient x = -17;
X 3 = -20 -60, (-20 + 3 = -17)
Step 3:
P = 3y ^ 2 – 17 and – 20 = 3y ^ 2 – 20y + 3y – 20
Step 4:
P = Y (3y – 20) + 1 (3y – 20)
Step 5:
P = (3y – 20) (y + 1)
Therefore, 3y ^ 2 – 17 and – 20 = (3y – 20) (y + 1) Response.
Example 5:
Factor 2 – 5y – 18Y ^ 2
Solution:
Let P = 2 – 5y – 18Y 18Y ^ 2 =- ^ 2 – 5y + 2
Now, follow the five steps listed above.
Step 1:
(Coefficient and ^ 2) x (constant term) = x 2 = -18 -36
Step 2:
We must express -36 as two factors whose sum is the coefficient of y = -5;
-36 = -2 = -2 X 18 X 2 x 4 x 9 = -9; [4 + (-9) = -5]
Step 3:
18Y =- P ^ 2 – 5y + 2 =- 18Y ^ 2 + 4y – 9y + 2
Step 4:
P = 2a ("9y + 2) + 1 (-9y + 2)
Step 5:
P = (-9y + 2) (2a + 1)
Thus, 2 – 5y – 18Y ^ 2 = (-9y + 2) (2a + 1) Response.
Example 6:
Factor (y ^ 2 + y) ^ 2 -18 (y ^ 2 + y) + 72
Solution:
Let P = (y ^ 2 + y) ^ 2 -18 (y ^ 2 + y) + 72
Put (y ^ 2 + y) = t, then P = t ^ 2-18t + 72
Now, follow the five steps listed above.
Step 1:
(Coefficient of t ^ 2) x (constant term) = 1 x 72 = 72
Step 2:
We have to express 72 as two factors whose sum is the coefficient of t = -18;
72 = 12 x 6 x = -12 -6, [(-12) + (-6) = -18]
Step 3:
P = t ^ 2-18t + 72 = ^ t 2 – 12t – 6t + 72
Step 4:
P = t (t – 12) – 6 (t – 12)
Step 5:
P = (t – 12) (t – 6)
But t = (y ^ 2 + y);
Therefore, P = (t – 12) (t – 6)
y ^ 2 + y – 12 = y ^ 2 + 4y – 3y – 12 = y (y + 4) – 3 (y + 4) = (y + 4) (y – 3)
y ^ 2 + y – 6 = y ^ 2 + 3y – 2y – 6 = and (y + 3) – 2 (y + 3) = (y + 3) (y – 2)
See how these two quadratic polynomials are factorised with the knowledge of the 5 steps.
You may have mastered the five steps of factoring by this time, to write directly like this.
Therefore,
P = (y ^ 2 + y) ^ 2 -18 (y ^ 2 + y) + 72
= (Y ^ 2 + y – 12) (y ^ 2 + y – 6)
= (Y + 4) (y – 3) (y + 3) (y – 2) American National Standard.
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