
How do you factor the given polynomial completely: $5{{x}^{2}}-13x+6$?
Answer
559.2k+ views
Hint: We start solving the problem by making the necessary arrangements in the given polynomial. We then take the common factors out by grouping the first two terms and next two terms together to proceed through the problem. We then check the common factor of the obtained result and then take it out which will give us the required factors of the given polynomial.
Complete step by step answer:
According to the problem, we are asked to factorize the given polynomial $5{{x}^{2}}-13x+6$ completely.
We have given the polynomial $5{{x}^{2}}-13x+6$ ---(1).
We know that $-13x=-10x-3x$. Let us use this in equation (1).
$\Rightarrow 5{{x}^{2}}-13x+6=5{{x}^{2}}-10x-3x+6$.
$\Rightarrow 5{{x}^{2}}-13x+6=\left( 5x\times x \right)+\left( 5x\times -2 \right)+\left( -3\times x \right)+\left( -3\times -2 \right)$ ---(2).
Let us now take the common factors out by grouping the first two terms and next two terms from equation (2).
$\Rightarrow 5{{x}^{2}}-13x+6=5x\times \left( x-2 \right)-3\times \left( x-2 \right)$ ---(3).
From equation (3), we can see that both the terms have a common factor $\left( x-2 \right)$. So, let us take it common.
$\Rightarrow 5{{x}^{2}}-13x+6=\left( 5x-3 \right)\left( x-2 \right)$.
So, we have got the factors of the given polynomial $5{{x}^{2}}-13x+6$ as $\left( 5x-3 \right)$ and $\left( x-2 \right)$.
$\therefore $ The factorization of the given polynomial $5{{x}^{2}}-13x+6$ is $\left( 5x-3 \right)\left( x-2 \right)$.
Note: We can also solve this problem by first finding the zeros of the given polynomial which will give the required factors. We should keep in mind that multiplying the obtained factors must give the same polynomial given in problem as result. We can also find the sum product of zeroes of the given polynomial. Similarly, we can expect problems to factorize the polynomial ${{x}^{2}}-2x+1$ completely.
Complete step by step answer:
According to the problem, we are asked to factorize the given polynomial $5{{x}^{2}}-13x+6$ completely.
We have given the polynomial $5{{x}^{2}}-13x+6$ ---(1).
We know that $-13x=-10x-3x$. Let us use this in equation (1).
$\Rightarrow 5{{x}^{2}}-13x+6=5{{x}^{2}}-10x-3x+6$.
$\Rightarrow 5{{x}^{2}}-13x+6=\left( 5x\times x \right)+\left( 5x\times -2 \right)+\left( -3\times x \right)+\left( -3\times -2 \right)$ ---(2).
Let us now take the common factors out by grouping the first two terms and next two terms from equation (2).
$\Rightarrow 5{{x}^{2}}-13x+6=5x\times \left( x-2 \right)-3\times \left( x-2 \right)$ ---(3).
From equation (3), we can see that both the terms have a common factor $\left( x-2 \right)$. So, let us take it common.
$\Rightarrow 5{{x}^{2}}-13x+6=\left( 5x-3 \right)\left( x-2 \right)$.
So, we have got the factors of the given polynomial $5{{x}^{2}}-13x+6$ as $\left( 5x-3 \right)$ and $\left( x-2 \right)$.
$\therefore $ The factorization of the given polynomial $5{{x}^{2}}-13x+6$ is $\left( 5x-3 \right)\left( x-2 \right)$.
Note: We can also solve this problem by first finding the zeros of the given polynomial which will give the required factors. We should keep in mind that multiplying the obtained factors must give the same polynomial given in problem as result. We can also find the sum product of zeroes of the given polynomial. Similarly, we can expect problems to factorize the polynomial ${{x}^{2}}-2x+1$ completely.
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