This simpler case uses the identity:
To factor a trinomial like $x^2 + Bx + C$, you need to find two numbers, let’s call them $a$ and $b$, such that:
Their product equals the constant term ($ab = C$).
Their sum equals the coefficient of the middle term ($a+b = B$).
Identify B and C: $B=5$ and $C=6$.
Find two numbers that multiply to 6 and add up to 5.
Factors of 6: (1, 6), (2, 3)
Sums: $1+6=7$, $2+3=5$. The numbers are 2 and 3.
Identify B and C: $B=-7$ and $C=12$.
Find two numbers that multiply to positive 12 and add up to negative 7. Since the product is positive and the sum is negative, both numbers must be negative.
Factors of 12: $(-1, -12)$, $(-2, -6)$, $(-3, -4)$
Sums: $-1+(-12)=-13$, $-2+(-6)=-8$, $-3+(-4)=-7$. The numbers are -3 and -4.
(Note: Factoring $-4$ out of $-4y+12$ results in the required $(y-3)$ binomial.)
When the leading coefficient ($a$) is not 1, the process is slightly different but still relies on splitting the middle term.
The steps are:
Calculate the product $A \times C$ (the coefficient of $x^2$ times the constant term).
Find two numbers ($p$ and $q$) that multiply to $A \times C$ and add up to $B$ (the middle coefficient).
Split the middle term $Bx$ into $px + qx$.
Factor by Grouping the resulting four terms.
Calculate $A \times C$: $3 \times 3 = 9$.
Find two numbers that multiply to 9 and add up to $B=10$.
Factors of 9: (1, 9), (3, 3)
Sums: $1+9=10$, $3+3=6$. The numbers are 1 and 9.
Goal: Factor the trinomial $4y^2 – 8y + 3$.
The product of the leading coefficient ($A=4$) and the constant term ($C=3$) is $4 \times 3 = 12$.
We need two numbers that:
Multiply to 12
Add up to the middle coefficient $-8$
The pairs that multiply to 12 are $1 \times 12$, $2 \times 6$, and $3 \times 4$. To get a positive product and a negative sum, both numbers must be negative. The correct pair is $-2$ and $-6$ (since $-2 + (-6) = -8$).
Replace $-8y$ with $-2y – 6y$:
Group the terms and find the common factors:
Goal: Factor the trinomial $2x^2 – 3x – 2$.
$A=2$ and $C=-2$. The product is $2 \times (-2) = -4$.
We need two numbers that:
Multiply to $-4$
Add up to $-3$
The factors of $4$ are $1$ and $4$, $2$ and $2$. To get a negative product, one number must be negative. The correct pair is $1$ and $-4$ (since $1 + (-4) = -3$).
Replace $-3x$ with $+x – 4x$:
Goal: Factor the trinomial $6x^2 + 5x – 6$.
$A=6$ and $C=-6$. The product is $6 \times (-6) = -36$.
We need two numbers that:
Multiply to $-36$
Add up to $5$
The correct pair is $9$ and $-4$ (since $9 + (-4) = 5$).
Replace $5x$ with $-4x + 9x$:
Goal: Factor the trinomial $3x^2 + 22x + 35$.
$A=3$ and $C=35$. The product is $3 \times 35 = 105$.
We need two numbers that:
Multiply to 105
Add up to $22$
By checking factors (e.g., $105 \div 5 = 21$), we find the correct pair is $7$ and $15$ (since $7 + 15 = 22$).
Replace $22x$ with $15x + 7x$:
Expression: $12x^2 – 17xy + 6y^2$
This is treated similarly, where the variable $y$ is attached to the terms in the split.
$A=12$ and $C=6$. The product is $12 \times 6 = 72$.
We need two numbers that:
Multiply to $72$
Add up to $-17$ (the coefficient of the $xy$ term)
To get a positive product and a negative sum, both numbers must be negative. The correct pair is $-8$ and $-9$ (since $-8 + (-9) = -17$).
Replace $-17xy$ with $-9xy – 8xy$:
Expression: $(2a-b)^2 + 2(2a-b) – 8$
This expression is complex, but we can simplify it by introducing a substitution for the repeated term $(2a-b)$.
Let $p = 2a – b$.
The expression becomes: $p^2 + 2p – 8$
We need two numbers that multiply to $-8$ and add up to $2$. These numbers are $4$ and $-2$.
Now, replace $p$ with $(2a-b)$.
Expression: $6 – x – 2x^2$
It’s usually easier to factor a quadratic when the terms are ordered by degree, and the leading coefficient is positive.
We look for two numbers that multiply to $(2 \times -6) = -12$ and add up to $1$ (the coefficient of $x$). These numbers are $4$ and $-3$.
To eliminate the negative sign outside, you can distribute it to one of the factors, for instance, $(2x-3)$:
Before diving into the next examples, it’s crucial to recall the formulas for squaring a trinomial:
Note:
(Notice how the negative terms $2bc$ and $2ca$ indicate that $c$ is the negative term in the squared binomial.)
Expression: $4a^2 + b^2 + c^2 + 4ab + 2bc + 4ac$
Rewrite the squared terms as perfect squares:
Step 2: Check the Cross-Product Terms
$2(2a)(b) = 4ab$ (Matches)
$2(b)(c) = 2bc$ (Matches)
$2(c)(2a) = 4ac$ (Matches)
The expression fits the pattern $(a+b+c)^2$, where the terms are $2a$, $b$, and $c$.
Expression: $9a^2 + 4b^2 + 16 + 12ab + 16b + 24a$
We have three squared terms: $9a^2 = (3a)^2$, $4b^2 = (2b)^2$, and $16 = (4)^2$.
Let $a\’=3a$, $b\’=2b$, and $c\’=4$.
$2(a\’)(b\’) = 2(3a)(2b) = 12ab$ (Matches)
$2(b\’)(c\’) = 2(2b)(4) = 16b$ (Matches)
$2(c\’)(a\’) = 2(4)(3a) = 24a$ (Matches)
Expression: $a^2 + b^2 + c^2 + 2ab – 4bc – 4ca$
We use the cross-product terms to determine which variable must be negative.
The $2ab$ term is positive, suggesting $a$ and $b$ have the same sign (let’s assume positive).
The $-4bc$ term is negative, suggesting one of $b$ or $c$ is negative. Since we assumed $b$ is positive, $c$ must be negative.
The $-4ca$ term is negative, suggesting one of $c$ or $a$ is negative. Since $a$ is positive, $c$ must be negative.
This confirms that the terms in the square are $a$, $b$, and $-c$.
Step 3: Check the Products with $-c$
$2(a)(b) = 2ab$ (Matches)
$2(b)(-c) = -2bc$ (Wait, the original is $-4bc$. Let’s re-examine the coefficients.)
Looking closely at the original expression: $a^2 + b^2 + \mathbf{4c^2} + 2ab – 4bc – 4ca$ (The $c^2$ term is likely $\mathbf{4c^2}$ based on the cross products $4bc$ and $4ca$). Let’s assume the correct expression has $4c^2$.
Assuming the corrected expression is: $a^2 + b^2 + \mathbf{4c^2} + 2ab – 4bc – 4ca$
New Squared Terms: $a^2$, $b^2$, and $(2c)^2$.
Cross-Products with $-2c$:
$2(a)(b) = 2ab$ (Matches)
$2(b)(-2c) = -4bc$ (Matches)
$2(a)(-2c) = -4ac$ (Matches)
The terms are $a$, $b$, and $-2c$.