
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 5th, 2017, 01:51 AM  #1 
Newbie Joined: Dec 2016 From: nottinghamshire,England Posts: 2 Thanks: 0  Quadratic by factorising
Hi All I need help with the method (not the answer) how do I got about solving 10x^2+3x4=0 I need to solve the following quadratic by factorising thank you 
January 5th, 2017, 02:20 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,519 Thanks: 506 Math Focus: Yet to find out. 
Many ways. You could start by dividing through by 10. $10x^2 + 3x  4 = 0$ Divide through by 10, $x^2 + \frac{3}{10} x  \frac{4}{10} = 0.$ Then ask yourself, what two numbers add together to give $\frac{3}{10}$, and multiply together to give $\frac{4}{10}$. With a bit of fiddling you should find that, $ \frac{1}{2} + \frac{4}{5} = \frac{3}{10}$ and that $ \frac{1}{2} * \frac{4}{5} =  \frac{4}{10}.$ Finally we have, $(x  \frac{1}{2})(x + \frac{4}{5}) = 0.$ And so, the solutions for $\displaystyle x$ are, $x = \frac{1}{2},$ and $x =  \frac{4}{5}.$ 
January 5th, 2017, 06:26 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,701 Thanks: 1359  
January 5th, 2017, 05:05 PM  #4  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
$\displaystyle \begin{align*} \left( x + m \right) \left( x + n \right) &= x \left( x + n \right) + m \left( x + n \right) \\ &= x^2 + n\,x + m\,x + m\,n \\ &= x^2 + \left( m + n \right) x + m\,n \end{align*}$ which is now a monic quadratic of the form $\displaystyle \begin{align*} x^2 + b\,x + c \end{align*}$, where $\displaystyle \begin{align*} b = m + n \end{align*}$ and $\displaystyle \begin{align*} c = m\,n \end{align*}$. So if you wanted to go backwards, you need to find the right two numbers (m and n) which multiply to give b and add to give c. Things get a little trickier when the quadratic is not monic though. Let's look at what happens when we expand $\displaystyle \begin{align*} \left( m\,x + n \right) \left( p\,x + q \right) \end{align*}$ (here m, n, p, q are just constants). It also helps to remember that any combination of them is also just a constant. Anyway, we get $\displaystyle \begin{align*} \left( m\,x + n \right) \left( p\,x + q \right) &= m\,x \left( p\,x + q \right) + n \left( p\,x + q \right) \\ &= m\,p\,x^2 + m\,q\,x + n\,p\,x + n\,q \\ &= m\,p\,x^2 + \left( m\,q + n\,p \right) x + n\,q \end{align*}$ which is now of the form $\displaystyle \begin{align*} a\,x^2 + b\,x + c \end{align*}$ where $\displaystyle \begin{align*} a = m\,p , \, b = m\,q + n\,p , \, c = n\,q \end{align*}$. Here it is not as simple as the monic case, because the middle term depends on four constants. But what we can do is multiply the a and c terms together. This gives $\displaystyle \begin{align*} a \cdot c &= m\,p \cdot n\,q \\ &= m\,q \cdot n\,p \end{align*}$ Now as I said before, any combination of the four constants is also just a constant. So here, when we have multiplied the a and c values together, it again gives the same combination of two numbers (mp and nq) multiplied together as the middle term does when they're added. So we can still use the "times to give, add to give" process, as long as we have first multiplied the a and c terms. Let's look at your example $\displaystyle \begin{align*} 10\,x^2 + 3\,x  4 \end{align*}$. We need to multiply a and c, which is $\displaystyle \begin{align*} 10 \left( 4 \right) = 40 \end{align*}$. So what we want are two numbers that multiply to give 40 (as they multiply to give a negative number, we know they need different signs) and add to give 3 (well really have a difference of 3 as we know that they have to have different signs  also the larger of the two numbers will be the positive). The two numbers are 8 and 5. So now we break the middle term up as the sum of these two numbers and continue to factorise... $\displaystyle \begin{align*} 10\,x^2 + 3\,x  4 &= 10\,x^2 + \left( 8  5 \right) x  4 \\ &= 10\,x^2 + 8\,x  5\,x  4 \\ &= 2\,x\left( 5\,x + 4 \right)  1 \left( 5\,x + 4 \right) \\ &= \left( 5\,x + 4 \right) \left( 2\,x  1 \right) \end{align*}$  
January 6th, 2017, 03:25 AM  #5 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
Use quadratic formula to factor out it, $x=\dfrac{b\pm \sqrt{b^24ac}}{2a}$ Put values in formula $\Rightarrow \dfrac{3\pm \sqrt{3^24×10(4)}}{2×10}$ $\Rightarrow \dfrac{3\pm \sqrt{169}}{20}$ $\Rightarrow \dfrac{3\pm 13}{20}$ $x=\dfrac{4}{5}$ $x=\dfrac{1}{2}$ 

Tags 
factorising, quadratic 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
factorising help  manich44  Algebra  10  December 12th, 2011 01:09 AM 
FACTORISING.... O_O  mathsilliterate  Algebra  1  September 2nd, 2010 03:50 AM 
Factorising quadratic equations  magician  Algebra  2  November 23rd, 2009 10:35 PM 
Factorising Please help  Dizzee  Algebra  5  October 21st, 2009 10:51 AM 
Factorising  Voltman  Algebra  2  April 12th, 2009 04:33 PM 