Question

factorise p^2+p-56 class 8 dav public school​

Answer 1

Given :-

$p {}^{2} + p - 56 = 0$

What to find out =Roots of the equation?

Solution :-

Factorise by splitting middle term

$p {}^{2} + p - 56 = 0 \\ \\ p {}^{2} + 8p - 7p - 56 = 0 \\ \\ p(p + 8) - 7(p + 8) = 0 \\ \\ (p + 8)(p - 7) = 0$

Now split it into possible cases

$(p + 8) = 0 \\ \\ (p - 7) = 0$

Hence,

$p_1 =( - 8) \\ \\ p_2 = (7)$

Extra information :-

• A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

$\sf p^2 + p - 56$

solve this by mid term split

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

$\sf\implies p^2 + p - 56$

$\sf\implies p^2 + 8p - 7p - 56$

$\sf\implies p(p + 8)-7(p+8)$

$\sf\implies(p+8)(p-7)$

roots in the equation

$\sf\implies p+8 = 0$

$\sf\implies p= - 8$

______________________

$\sf\implies p - 7 = 0$

$\sf\implies p = 7$

${\bold{\blue{\boxed{\bf{p_1 = -8}}}}}$ And ${\bold{\blue{\boxed{\bf{p_2 = 7 }}}}}$

__________________________❤

THANK YOU❤