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Let's start by finding the GCF of 4 and 12. Factors of4:& 1,2,and 4 Factors of12:& 1,2,3, 4,6,and 12 We found that the GCF of the coefficients is 4. To find the GCF of the variables, we need to identify the variables repeated in both terms and write them with their minimum exponents. \begin{aligned} \textbf{Factors of }\bm{1^\text{st}}\textbf{ Variable:}&\ {\color{#FF0000}{x}}, x^2\\ \textbf{Factors of }\bm{2^\text{nd}}\textbf{ Variable:}&\ {\color{#FF0000}{x}} \end{aligned} We see that there is one repeated variable factor, x. Thus, the GCF of the expression is 4* x= 4x. Now we can write the given expression in terms of the GCF. 4x^2-12x ⇔ 4x* x- 4x* 3 Finally, we will factor out the GCF. 4x* x- 4x* 3 ⇔ 4x(x-3)
The result of factoring out a GCF from the given expression is a trinomial with a leading coefficient of 1. 3(y^2+2y+1) Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.
We want to factor the above expression. Factoring is much easier when our polynomial is a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to ask ourselves three questions.
Is the first term a perfect square? | y^2= y^2 âś“ |
Is the last term a perfect square? | 1= 1^2 âś“ |
Is the middle term twice the product of 1 and y? | 2y=2* 1* y âś“ |
As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is an addition sign in the middle. y^2+2y+1 ⇔ ( y+ 1)^2 Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 3(y+1)^2
Split into factors
Factor out m
Here we have a quadratic trinomial of the form am^2+bm+c, where |a| ≠1 and there are no common factors. To factor this expression, we will rewrite the middle term, bm, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. m( 2m^2+7m+3 ) We have that a= 2, b=7, and c=3. There are now three steps we need to follow in order to rewrite the above expression.
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 & 6 & 1 + 6 &7 2 &3 &2 + 3 &5
c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &12 &- 1 + 12 &11 - 2 & 6 & - 2 + 6 &4 - 3 &4 &- 3 + 4 &1