NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 29 Exercise 14.2

Factorise the expression:

p^{2}+6 p+8



A polynomial called a "quadratic" is expressed as "ax2 + bx + c," where "a," "b," and "c" are simple numbers. We may get the two integers that will multiply to equal the product term "ac" and add up to equal "b," the coefficient on the x-term.

We observed that, 8 = 4 x 2 and 4 + 2 = 6

\begin{aligned} &\mathrm{p}^{2}+6 \mathrm{p}+8 \text { can be written as } \mathrm{p}^{2}+2 \mathrm{p}+4 \mathrm{p}+8\\ &\text { Taking Common terms, we get } \end{aligned}

\begin{aligned} &p^{2}+6 p+8=p^{2}+2 p+4 p+8=p(p+2)+4(p+2)\\ &\text { Again } p+2 \text { is common in both the terms. } \end{aligned}

\begin{array}{l} =(p+2)(p+4) \\ \text { This implies: } p^{2}+6 p+8=(p+2)(p+4) \end{array}

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