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Question

Question: How do you factor \[y={{x}^{2}}-7x+10\]?...

How do you factor y=x27x+10y={{x}^{2}}-7x+10?

Explanation

Solution

For answering this question we will use factorization. Factorization is the process of deriving factors of a number which divides the given number evenly. For this question we will split the constant term and we will use the sum product pattern of splitting and then we will simplify the equation using the x2a2=(x+a)(xa)\Rightarrow {{x}^{2}}-{{a}^{2}}=\left( x+a \right)\left( x-a \right) formula and simplify until we get the solution.

Complete step by step solution:
Now considering from the question we have an expression x27x+10{{x}^{2}}-7x+10 for which we need to derive the factors.
We can factor the x27x+10{{x}^{2}}-7x+10 by below method:
Given equation is in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0.
First we have to divide the constant term in the equation which is 1010 into the product of the two numbers and must make sure that the sum of the two numbers must be equal to the coefficient ofxx.
Now, the constant term 1010 can be split into the product of the two numbers in two ways those are10×1,5×210\times 1,5\times 2.
But here we have to take the splitting as 5×25\times 2 as the sum of 5 and 2 must be equal to the coefficient of xx.
So, 1010can be split into products of 22and 55.
Their sum is also equal to 77which is equal to the coefficient of xx.
So, the given question can be factored as follows.
x27x+10\Rightarrow {{x}^{2}}-7x+10
x25x2x+10\Rightarrow {{x}^{2}}-5x-2x+10
x(x5)2(x5)\Rightarrow x\left( x-5 \right)-2\left( x-5 \right)
(x2)(x5)\Rightarrow \left( x-2 \right)\left( x-5 \right)
Therefore, the factors will be (x2),(x5)\left( x-2 \right),\left( x-5 \right).

Note: During answering questions of this type we should be sure with our calculations. We can also use the formulae for obtaining the roots of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 and solve the question. So, the roots of the quadratic equation given as b±b24ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} then if the two solutions are p,qp,q then the factors will be (xp)(xq)\left( x-p \right)\left( x-q \right) . For x27x+10 {{x}^{2}}-7x+10 the roots are 7±494(10)2=7±32=5,2\dfrac{7\pm \sqrt{49-4\left( 10 \right)}}{2}=\dfrac{7\pm 3}{2}=5,2 then the factors will be (x2),(x5) \left( x-2 \right),\left( x-5 \right).