MHB Doubt in factoring a trinomial

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To factor the trinomial 3x² + 11x + 10, the expression can be rewritten as 3x² + 5x + 6x + 10. The key is to find two factors of the product of the leading coefficient and the constant term, which is 30, that add up to 11; these factors are 6 and 5. This leads to the factorization (3x + 5)(x + 2). An alternative method involves grouping, resulting in the same factorization. Understanding these steps is crucial for correctly factoring similar quadratic expressions.
mathlearn
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Hey (Wave) (Party),

Problem

Factor $3x^2+11x+10$

Workings

This expression can be separated into,

$3x^2+10x+x+10$

Where have i done wrong ? (Thinking)

Many Thanks :)
 
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What I would do here is observe that:

$$3\cdot10=30$$

$$6\cdot5=30$$

$$6+5=11$$

Hence:

$$3x^2+11x+10=(3x+5)(x+2)$$
 
MarkFL said:
What I would do here is observe that:

$$3\cdot10=30$$

$$6\cdot5=30$$

$$6+5=11$$

Hence:

$$3x^2+11x+10=(3x+5)(x+2)$$

Did you multiply the expression by 3 (Happy) to get 30.
 
mathlearn said:
Did you multiply the expression by 3 (Happy) to get 30.

I multiplied the first coefficient by the last to get 30, and then looked for two factors of this product whose sum is 11, which are 6 and 5.

Here's a tutorial I wrote on the subject:

http://mathhelpboards.com/math-notes-49/factoring-quadratics-3396.html
 
Alternatively,

$$\begin{align*}3x^2+11x+10&=3x^2+5x+6x+10 \\
&=x(3x+5)+2(3x+5) \\
&=(3x+5)(x+2)\end{align*}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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