# How to factor this expression?

#### ace7269

##### New member
I understand how to do basic factoring like ax^2 + bx + c but I don't understand how to do factoring with more variables then those in the equation. For example: [FONT=MathJax_Math]x[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]14[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]49[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]y
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#### JeffM

##### Elite Member
I understand how to do basic factoring like ax^2 + bx + c but I don't understand how to do factoring with more variables then those in the equation. For example: [FONT=MathJax_Math]x[FONT=MathJax_Main]2[/FONT][/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]14[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]49[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]y
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Why don't you provide an actual problem in its entirety and show what you have tried.

By the way, you have given expressions, not equations. You factor expressions.

And it is quite feasible to factor the given expression: $$\displaystyle x^2 + 14x + 49 - y = (x + 7)^2 - y.$$

#### ace7269

##### New member
Factoring

Okay, here is a full problem. Factor the expression [FONT=MathJax_Math]a[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]6[/FONT][/FONT]. Simplify your answer as much as possible.

Please trust me I have tried to work these out but I can't figure it out. Should I be using the ax^2 + bx + c expression? I just don't understand where to even start with these expressions. I'm not asking for anyone to do my homework I'm just trying to figure out these steps so I can do these kinds of problems on the exam tomorrow.

#### JeffM

##### Elite Member
Okay, here is a full problem. Factor the expression [FONT=MathJax_Math]a[FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]6[/FONT][/FONT]. Simplify your answer as much as possible.

Please trust me I have tried to work these out but I can't figure it out. Should I be using the ax^2 + bx + c expression? I just don't understand where to even start with these expressions. I'm not asking for anyone to do my homework I'm just trying to figure out these steps so I can do these kinds of problems on the exam tomorrow.
$$\displaystyle ax^2 + bx + c$$ is the standard form for a quadratic in one variable.

A quadratic in standard form can be factored using the quadratic formula, but it pertains only to that special case.

All factoring means is to restate an expression as a product of expressions. The most common way to do that is to look for common factors in the expression.

Is your expression $$\displaystyle ay + a^6y^6?$$ If so, you should see that ay is a common factor and $$\displaystyle ay + a^6y^6 = ay(1 + a^5y^5).$$

If your expression is $$\displaystyle ay + a * 6 * y * 6,$$ then ay is again a common factor and $$\displaystyle ay - a * 6 * y * 6 = ay(1 + 6 * 6) = 37ay.$$

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#### srmichael

##### Full Member
$$\displaystyle ax^2 + bx + c$$ is the standard form for a quadratic in one variable.

A quadratic in standard form can be factored using the quadratic formula, but it pertains only to that special case.

All factoring means is to restate an expression as a product of expressions. The most common way to do that is to look for common factors in the expression.

Is your expression $$\displaystyle ay + a^6y^6?$$ If so, you should see that ay is a common factor and $$\displaystyle ay + a^6y^6 = ay(1 + a^5y^5).$$

If your expression is $$\displaystyle ay - a * 6 * y * 6,$$ then ay is again a common factor and $$\displaystyle ay - a * 6 * y * 6 = ay(1 + 6 * 6) = 37ay.$$
JeffM, small typo. I believe you meant ay(1 - 6 * 6) = -35ay #### JeffM

##### Elite Member
JeffM, small typo. I believe you meant ay(1 - 6 * 6) = -35ay I had a typo all right: original expression required plus sign not a minus sign. Thanks