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January 5th, 2017, 12:51 AM   #1
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Question Quadratic by factorising

Hi All

I need help with the method (not the answer)

how do I got about solving


I need to solve the following quadratic by factorising

thank you
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January 5th, 2017, 01:20 AM   #2
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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,

$x = \frac{1}{2},$


$x = - \frac{4}{5}.$
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January 5th, 2017, 05:26 AM   #3
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Some good advice here ...

Factoring Trinomials When the Leading Coefficient is not 1
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January 5th, 2017, 04:05 PM   #4
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Originally Posted by claire123 View Post
Hi All

I need help with the method (not the answer)

how do I got about solving


I need to solve the following quadratic by factorising

thank you
Let's start by expanding out $\displaystyle \begin{align*} \left( x + m \right) \left( x + n \right) \end{align*}$, where m and n are just constants. Then we get

$\displaystyle \begin{align*} \left( x + m \right) \left( x + n \right) &= x \left( x + n \right) + m \left( x + n \right) \\ &= x^2 + n\,x + m\,x + m\,n \\ &= x^2 + \left( m + n \right) x + m\,n \end{align*}$

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...

$\displaystyle \begin{align*} 10\,x^2 + 3\,x - 4 &= 10\,x^2 + \left( 8 - 5 \right) x - 4 \\ &= 10\,x^2 + 8\,x - 5\,x - 4 \\ &= 2\,x\left( 5\,x + 4 \right) - 1 \left( 5\,x + 4 \right) \\ &= \left( 5\,x + 4 \right) \left( 2\,x - 1 \right) \end{align*}$
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January 6th, 2017, 02:25 AM   #5
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Use quadratic formula to factor out it,

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Put values in formula

$\Rightarrow \dfrac{-3\pm \sqrt{3^2-4×10(-4)}}{2×10}$

$\Rightarrow \dfrac{-3\pm \sqrt{169}}{20}$

$\Rightarrow \dfrac{-3\pm 13}{20}$


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