Question
$p=-2$, find the value of $-2 p^3-3 p^2+4 p+7$
✓
Answer
$=-2 p^3-3 p^2+4 p+7$
$=-2 p^3-3 p^2+4 p+7$
$=-2(-2)^3-3(-2)^2+4(-2)+7$
$=-2(-8)-3(4)-8+7=16-12-8+7$
$=16+7-12-8=23-20=3$
$=-2 p^3-3 p^2+4 p+7$
$=-2(-2)^3-3(-2)^2+4(-2)+7$
$=-2(-8)-3(4)-8+7=16-12-8+7$
$=16+7-12-8=23-20=3$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Add the following expression:
$5 x-2 x^2-8,8 x^2-7 x-9,3+7 x^2-2 x$
→$5 x-2 x^2-8,8 x^2-7 x-9,3+7 x^2-2 x$
Here, you find four nets. There are two correct nets among them to make a tetrahedron. See If you can work out which nets will make a tetrahedron?
Anamika thought of a number. She multiplied it by $2,$ added 5 to the product and obtained $17$ as the result. What is the number she had thought of $?$
→Multiply: $\frac{5}{7}\text{ by }\Big(\frac{-3}{4}\Big)$
→List five rational numbers between $\frac{-4}{5} \text { and } \frac{-2}{3}$
→Find: $128.9$ $\div$ $1000$
→Social Studies Application: Remembering that $1AD$ came immediately after $1BC,$ while solving these problems take $1BC$ as $-1$ and $1AD$ as $+1.$ Turks ruled Egypt in the year $1517AD$ and Queen Nefertis ruled Egypt about $2900$ years before the Turks ruled. In what year did she rule$?$
→In right $\triangle \text{ABC},$ the lengths of the legs are given. Find the length of the hypotenuse.
$a = 8\ cm, b = 15\ cm$
→$a = 8\ cm, b = 15\ cm$
Add the following expressions:
$x^3-2 x^2 y+3 x y^2-y^3, 2 x^3-5 x y^2+3 x^2 y-4 y^3$
→$x^3-2 x^2 y+3 x y^2-y^3, 2 x^3-5 x y^2+3 x^2 y-4 y^3$
Express following as a product of powers of their prime factors: $800$
→