My Math Forum Factoring

 Algebra Pre-Algebra and Basic Algebra Math Forum

 October 21st, 2017, 10:34 PM #1 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 Factoring A yacht sails a leg of a race by going in a straight course from point A to point B for (x^2 + 8x) km. From point B it travels to point C for (2x+11)km, then back to point A for (4x+24)km. a) write a simplified expression in km for the total distance the yacht sailed b) factorise the expression in (a) c) the factorised expression also represents the product of the length and width of a rectangle, and if the length is 15km, find the value of x. Give your answer as an exact distance d) Find the length of the race, rounded to two decimal places, by substituting the exact value for x into the expression for the total distance in (a) My answer: a) (x^2 + 8x) + (2x+11) + (4x+24) = x^2 + 14x + 35 b) I can't work out how to factorise (x^2 + 14x + 35). Any help would be appreciated.
 October 21st, 2017, 10:53 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 b) you can always use the quadratic formula \begin{align*} r_1, r_2 &= \dfrac{-14 \pm \sqrt{14^2 - (4)(35)}}{2} \\ &= \dfrac{-14 \pm \sqrt{56}}{2} \\ &= -7 \pm \sqrt{14} \end{align*} so $x^2 + 14x + 35 = (x-r_1)(x-r_2) = (x + 7 - \sqrt{14})(x + 7 + \sqrt{14})$ Thanks from pianist
 October 21st, 2017, 11:10 PM #3 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
October 21st, 2017, 11:20 PM   #4
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,100
Thanks: 1093

Quote:
 Originally Posted by pianist Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
I suppose if you are really good you can "foil" the thing by eye even though the roots aren't rational.

The quadratic formula is the way to go when the factorization isn't immediately apparent.

October 23rd, 2017, 10:33 AM   #5
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Quote:
 Originally Posted by pianist Thanks, so in this case is there another way to solve it or is the quadratic formula the only way?
You get the "quadratic formula" itself by applying "completing the square" to the general quadratic $\displaystyle ax^2+ bx+ c= 0$.

Here, the quadratic expression is $\displaystyle x^2+ 14x+ 35$. Knowing that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$ and comparing those two, 2a= 14 so a= 7, $\displaystyle a^2= 49$. $\displaystyle x^2+ 14x+ 35= x^2+ 14x+ 49- 49+ 35= (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2= (x+ 7+ \sqrt{14})(x+ 7- \sqrt{14})$.

Last edited by skipjack; October 24th, 2017 at 05:12 AM.

October 24th, 2017, 04:02 AM   #6
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

Quote:
 Originally Posted by Country Boy You get the "quadratic formula" itself by applying "completing the square" to the general quadratic $\displaystyle ax^2+ bx+ c= 0$. Here, the quadratic expression is $\displaystyle x^2+ 14x+ 35$. Knowing that $\displaystyle (x+ a)^2= x^2+ 2ax+ a^2$ and comparing those two, 2a= 14 so a= 7, $\displaystyle a^2= 49$. $\displaystyle x^2+ 14x+ 35= x^2+ 14x+ 49- 49+ 35= (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2= (x+ 7+ \sqrt{14})(x+ 7- \sqrt{14})$.
Two typos in this. First, "$\displaystyle a^2= 49$" should have been "$\displaystyle a^2= 4$".

More importantly, "$\displaystyle (x+ 7)^2- 14= (x+ 7)^2+ (\sqrt{14})^2$" should have been "$\displaystyle (x+ 7)^2- 14= (x+ 7)^2- (\sqrt{14})^2$". Then we can use "$\displaystyle a^2- b^2= (a+ b)(a- b)$.

Last edited by skipjack; October 24th, 2017 at 05:13 AM.

 Tags factoring

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post vagulus Elementary Math 19 February 15th, 2015 09:22 PM vagulus Elementary Math 1 January 25th, 2015 08:03 PM Joe C Applied Math 4 January 31st, 2014 12:21 PM dunn Algebra 4 September 28th, 2011 10:40 PM hannah2329 Algebra 3 September 13th, 2011 04:20 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top